Expected value

"E(X)" redirects here. For the ${\displaystyle e^{x}}$ function, see Exponential function.

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In probability theory, the expected value of a random variable ${\displaystyle X}$, often denoted ${\displaystyle \operatorname {E} (X)}$, ${\displaystyle \operatorname {E} [X]}$, or ${\displaystyle EX}$, is a generalization of the weighted average, and is intuitively the arithmetic mean of a large number of independent realizations of ${\displaystyle X}$. The expectation operator ${\displaystyle \operatorname {E} }$ is also commonly stylized as ${\displaystyle E}$ or ${\displaystyle \mathbb {E} }$. ^[1][2][3][4] The expected value is also known as the expectation, mathematical expectation, mean, average, or first moment. Expected value is a key concept in economics, finance, and many other subjects.

By definition, the expected value of a constant random variable ${\displaystyle X=c}$ is ${\displaystyle c}$.^[5] The expected value of a random variable ${\displaystyle X}$ with equiprobable outcomes ${\displaystyle \{c_{1},\ldots ,c_{n}\}}$ is defined as the arithmetic mean of the terms ${\displaystyle c_{i}.}$ If some of the probabilities ${\displaystyle \Pr \,(X=c_{i})}$ of an individual outcome ${\displaystyle c_{i}}$ are unequal, then the expected value is defined to be the probability-weighted average of the ${\displaystyle c_{i}}$s, that is, the sum of the ${\displaystyle n}$ products ${\displaystyle c_{i}\cdot \Pr \,(X=c_{i})}$.^[6] The expected value of a general random variable involves integration in the sense of Lebesgue.

History[]

The idea of the expected value originated in the middle of the 17th century from the study of the so-called problem of points, which seeks to divide the stakes in a fair way between two players, who have to end their game before it is properly finished.^[7] This problem had been debated for centuries, and many conflicting proposals and solutions had been suggested over the years, when it was posed to Blaise Pascal by French writer and amateur mathematician Chevalier de Méré in 1654. Méré claimed that this problem couldn't be solved, and that it showed just how flawed mathematics was when it came to its application to the real world. Pascal, being a mathematician, was provoked and determined to solve the problem once and for all.

He began to discuss the problem in the famous series of letters to Pierre de Fermat. Soon enough, they both independently came up with a solution. They solved the problem in different computational ways, but their results were identical because their computations were based on the same fundamental principle. The principle is that the value of a future gain should be directly proportional to the chance of getting it. This principle seemed to have come naturally to both of them. They were very pleased by the fact that they had found essentially the same solution, and this in turn made them absolutely convinced that they had solved the problem conclusively; however, they did not publish their findings. They only informed a small circle of mutual scientific friends in Paris about it.^[8]

Three years later, in 1657, a Dutch mathematician Christiaan Huygens, who had just visited Paris, published a treatise (see Huygens (1657)) "De ratiociniis in ludo aleæ" on probability theory. In this book, he considered the problem of points, and presented a solution based on the same principle as the solutions of Pascal and Fermat. Huygens also extended the concept of expectation by adding rules for how to calculate expectations in more complicated situations than the original problem (e.g., for three or more players). In this sense, this book can be seen as the first successful attempt at laying down the foundations of the theory of probability.

In the foreword to his book, Huygens wrote:

It should be said, also, that for some time some of the best mathematicians of France have occupied themselves with this kind of calculus so that no one should attribute to me the honour of the first invention. This does not belong to me. But these savants, although they put each other to the test by proposing to each other many questions difficult to solve, have hidden their methods. I have had therefore to examine and go deeply for myself into this matter by beginning with the elements, and it is impossible for me for this reason to affirm that I have even started from the same principle. But finally I have found that my answers in many cases do not differ from theirs.

— Edwards (2002)

Thus, Huygens learned about de Méré's Problem in 1655 during his visit to France; later on in 1656 from his correspondence with Carcavi, he learned that his method was essentially the same as Pascal's; so that before his book went to press in 1657, he knew about Pascal's priority in this subject.

In the mid-nineteenth century, Pafnuty Chebyshev became the first person to think systematically in terms of the expectations of random variables.^[9]

Etymology[]

Neither Pascal nor Huygens used the term "expectation" in its modern sense. In particular, Huygens writes:^[10]

That any one Chance or Expectation to win any thing is worth just such a Sum, as wou'd procure in the same Chance and Expectation at a fair Lay. ... If I expect a or b, and have an equal chance of gaining them, my Expectation is worth (a+b)/2.

More than a hundred years later, in 1814, Pierre-Simon Laplace published his tract "Théorie analytique des probabilités", where the concept of expected value was defined explicitly:^[11]

… this advantage in the theory of chance is the product of the sum hoped for by the probability of obtaining it; it is the partial sum which ought to result when we do not wish to run the risks of the event in supposing that the division is made proportional to the probabilities. This division is the only equitable one when all strange circumstances are eliminated; because an equal degree of probability gives an equal right for the sum hoped for. We will call this advantage mathematical hope.

Notations[]

The use of the letter ${\displaystyle \mathop {\hbox{E}} }$ to denote expected value goes back to W. A. Whitworth in 1901.^[12] The symbol has become popular since then for English writers. In German, ${\displaystyle \mathop {\hbox{E}} }$ stands for "Erwartungswert", in Spanish for "Esperanza matemática", and in French for "Espérance mathématique".^[13]

When E is used to denote expected value, authors use a variety of notation: the expectation operator can be styalized as ${\displaystyle \operatorname {E} }$ (upright), ${\displaystyle E}$ (italic), or ${\displaystyle \mathbb {E} }$ (in blackboard bold), while round brackets (${\displaystyle E(X)}$), square brackets (${\displaystyle E[X]}$), or no brackets (${\displaystyle EX}$) are all used.

Another popular notation is ${\displaystyle \mu _{X}}$, whereas ${\displaystyle \langle X\rangle }$ is commonly used in physics, and ${\displaystyle \mathop {\hbox{M}} (X)}$ in Russian-language literature.

Definition[]

Finite case[]

Let ${\displaystyle X}$ be a (discrete) random variable with a finite number of finite outcomes ${\displaystyle x_{1},x_{2},\ldots ,x_{k}}$ occurring with probabilities ${\displaystyle p_{1},p_{2},\ldots ,p_{k},}$ respectively. The expectation of ${\displaystyle X}$ is defined as ^[6]

${\displaystyle \operatorname {E} [X]=\sum _{i=1}^{k}x_{i}\,p_{i}=x_{1}p_{1}+x_{2}p_{2}+\cdots +x_{k}p_{k}.}$

Since ${\displaystyle p_{1}+p_{2}+\cdots +p_{k}=1,}$ the expected value is the weighted sum of the ${\displaystyle x_{i}}$ values, with the probabilities ${\displaystyle p_{i}}$ as the weights.

If all outcomes ${\displaystyle x_{i}}$ are equiprobable (that is, ${\displaystyle p_{1}=p_{2}=\cdots =p_{k}}$), then the weighted average turns into the simple average. On the other hand, if the outcomes ${\displaystyle x_{i}}$ are not equiprobable, then the simple average must be replaced with the weighted average, which takes into account the fact that some outcomes are more likely than others.

An illustration of the convergence of sequence averages of rolls of a die to the expected value of 3.5 as the number of rolls (trials) grows.

Examples[]

• Let ${\displaystyle X}$ represent the outcome of a roll of a fair six-sided die. More specifically, ${\displaystyle X}$ will be the number of pips showing on the top face of the die after the toss. The possible values for ${\displaystyle X}$ are 1, 2, 3, 4, 5, and 6, all of which are equally likely with a probability of 1/6. The expectation of ${\displaystyle X}$ is

${\displaystyle \operatorname {E} [X]=1\cdot {\frac {1}{6}}+2\cdot {\frac {1}{6}}+3\cdot {\frac {1}{6}}+4\cdot {\frac {1}{6}}+5\cdot {\frac {1}{6}}+6\cdot {\frac {1}{6}}=3.5.}$

If one rolls the die ${\displaystyle n}$ times and computes the average (arithmetic mean) of the results, then as ${\displaystyle n}$ grows, the average will almost surely converge to the expected value, a fact known as the strong law of large numbers.

• The roulette game consists of a small ball and a wheel with 38 numbered pockets around the edge. As the wheel is spun, the ball bounces around randomly until it settles down in one of the pockets. Suppose random variable ${\displaystyle X}$ represents the (monetary) outcome of a $1 bet on a single number ("straight up" bet). If the bet wins (which happens with probability 1/38 in American roulette), the payoff is $35; otherwise the player loses the bet. The expected profit from such a bet will be

${\displaystyle \operatorname {E} [\,{\text{gain from }}\$1{\text{ bet}}\,]=-\$1\cdot {\frac {37}{38}}+\$35\cdot {\frac {1}{38}}=-\${\frac {1}{19}}.}$

That is, the bet of $1 stands to lose ${\displaystyle -\${\frac {1}{19}}}$, so its expected value is ${\displaystyle -\${\frac {1}{19}}.}$

Countably infinite case[]

Intuitively, the expectation of a random variable taking values in a countable set of outcomes is defined analogously as the weighted sum of the outcome values, where the weights correspond to the probabilities of realizing that value. However, convergence issues associated with the infinite sum necessitate a more careful definition. A rigorous definition first defines expectation of a non-negative random variable, and then adapts it to general random variables.

Let ${\displaystyle X}$ be a non-negative random variable with a countable set of outcomes ${\displaystyle x_{1},x_{2},\ldots ,}$ occurring with probabilities ${\displaystyle p_{1},p_{2},\ldots ,}$ respectively. Analogous to the discrete case, the expected value of ${\displaystyle X}$ is then defined as the series

${\displaystyle \operatorname {E} [X]=\sum _{i=1}^{\infty }x_{i}\,p_{i}.}$

Note that since ${\displaystyle x_{i}p_{i}\geq 0}$, the infinite sum is well-defined and does not depend on the order in which it is computed. Unlike the finite case, the expectation here can be equal to infinity, if the infinite sum above increases without bound.

For a general (not necessarily non-negative) random variable ${\displaystyle X}$ with a countable number of outcomes, set ${\displaystyle X^{+}(\omega )=\max(X(\omega ),0)}$ and ${\displaystyle X^{-}(\omega )=-\min(X(\omega ),0)}$. By definition,

${\displaystyle \operatorname {E} [X]=\operatorname {E} [X^{+}]-\operatorname {E} [X^{-}].}$

Like with non-negative random variables, ${\displaystyle \operatorname {E} [X]}$ can, once again, be finite or infinite. The third option here is that ${\displaystyle \operatorname {E} [X]}$ is no longer guaranteed to be well defined at all. The latter happens whenever ${\displaystyle \operatorname {E} [X^{+}]=\operatorname {E} [X^{-}]=\infty }$.

Examples[]

• Suppose ${\displaystyle x_{i}=i}$ and ${\displaystyle p_{i}={\frac {k}{i2^{i}}},}$ for ${\displaystyle i=1,2,3,\ldots }$, where ${\displaystyle k={\frac {1}{\ln 2}}}$ (with ${\displaystyle \ln }$ being the natural logarithm) is the scale factor such that the probabilities sum to 1. Then, using the direct definition for non-negative random variables, we have

${\displaystyle \operatorname {E} [X]=\sum _{i}x_{i}p_{i}=1\left({\frac {k}{2}}\right)+2\left({\frac {k}{8}}\right)+3\left({\frac {k}{24}}\right)+\dots ={\frac {k}{2}}+{\frac {k}{4}}+{\frac {k}{8}}+\dots =k.}$

• An example where the expectation is infinite arises in the context of the St. Petersburg paradox. Let ${\displaystyle x_{i}=2^{i}}$ and ${\displaystyle p_{i}={\frac {1}{2^{i}}}}$ for ${\displaystyle i=1,2,3,\ldots }$. Once again, since the random variable is non-negative, the expected value calculation gives

${\displaystyle \operatorname {E} [X]=\sum _{i=1}^{\infty }x_{i}\,p_{i}=2\cdot {\frac {1}{2}}+4\cdot {\frac {1}{4}}+8\cdot {\frac {1}{8}}+16\cdot {\frac {1}{16}}+\cdots =1+1+1+1+\cdots \,=\infty .}$

• For an example where the expectation is not well-defined, suppose the random variable ${\displaystyle X}$ takes values ${\displaystyle k=1,-2,3,-4,\cdots }$ with respective probabilities ${\displaystyle {\frac {c}{1^{2}}},{\frac {c}{2^{2}}},{\frac {c}{3^{2}}},{\frac {c}{4^{2}}}}$, ..., where ${\displaystyle c={\frac {6}{\pi ^{2}}}}$ is a normalizing constant that ensures the probabilities sum up to one.

Then it follows that ${\displaystyle X^{+}}$ takes value ${\displaystyle (2k-1)}$ with probability ${\displaystyle c/(2k-1)^{2}}$ for ${\displaystyle k=1,2,3,\cdots }$ and takes value ${\displaystyle 0}$ with remaining probability. Similarly, ${\displaystyle X^{-}}$ takes value ${\displaystyle 2k}$ with probability ${\displaystyle c/(2k)^{2}}$ for ${\displaystyle k=1,2,3,\cdots }$ and takes value ${\displaystyle 0}$ with remaining probability. Using the definition for non-negative random variables, one can show that both ${\displaystyle \operatorname {E} [X^{+}]=\infty }$ and ${\displaystyle \operatorname {E} [X^{-}]=\infty }$ (see Harmonic series). Hence, the expectation of ${\displaystyle X}$ is not well-defined.

Absolutely continuous case[]

If ${\displaystyle X}$ is a random variable with a probability density function of ${\displaystyle f(x)}$, then the expected value is defined as the Lebesgue integral

${\displaystyle \operatorname {E} [X]=\int _{\mathbb {R} }xf(x)\,dx,}$

where the values on both sides are well defined or not well defined simultaneously.

Example. A random variable that has the Cauchy distribution^[14] has a density function, but the expected value is undefined since the distribution has large "tails".

General case[]

In general, if ${\displaystyle X}$ is a random variable defined on a probability space ${\displaystyle (\Omega ,\Sigma ,\operatorname {P} )}$, then the expected value of ${\displaystyle X}$, denoted by ${\displaystyle \operatorname {E} [X]}$, is defined as the Lebesgue integral

${\displaystyle \operatorname {E} [X]=\int _{\Omega }X(\omega )\,d\operatorname {P} (\omega ).}$

For multidimensional random variables, their expected value is defined per component. That is,

${\displaystyle \operatorname {E} [(X_{1},\ldots ,X_{n})]=(\operatorname {E} [X_{1}],\ldots ,\operatorname {E} [X_{n}])}$

and, for a random matrix ${\displaystyle X}$ with elements ${\displaystyle X_{ij}}$, ${\displaystyle (\operatorname {E} [X])_{ij}=\operatorname {E} [X_{ij}].}$

Basic properties[]

The basic properties below (and their names in bold) replicate or follow immediately from those of Lebesgue integral. Note that the letters "a.s." stand for "almost surely"—a central property of the Lebesgue integral. Basically, one says that an inequality like ${\displaystyle X\geq 0}$ is true almost surely, when the probability measure attributes zero-mass to the complementary event ${\displaystyle \left\{X<0\right\}}$.

• For a general random variable ${\displaystyle X}$, define as before ${\displaystyle X^{+}(\omega )=\max(X(\omega ),0)}$ and ${\displaystyle X^{-}(\omega )=-\min(X(\omega ),0)}$, and note that ${\displaystyle X=X^{+}-X^{-}}$, with both ${\displaystyle X^{+}}$ and ${\displaystyle X^{-}}$ nonnegative, then:

${\displaystyle \operatorname {E} [X]={\begin{cases}\operatorname {E} [X^{+}]-\operatorname {E} [X^{-}]&{\text{if }}\operatorname {E} [X^{+}]<\infty {\text{ and }}\operatorname {E} [X^{-}]<\infty ;\\\infty &{\text{if }}\operatorname {E} [X^{+}]=\infty {\text{ and }}\operatorname {E} [X^{-}]<\infty ;\\-\infty &{\text{if }}\operatorname {E} [X^{+}]<\infty {\text{ and }}\operatorname {E} [X^{-}]=\infty ;\\{\text{undefined}}&{\text{if }}\operatorname {E} [X^{+}]=\infty {\text{ and }}\operatorname {E} [X^{-}]=\infty .\end{cases}}}$

• Let ${\displaystyle {\mathbf {1} }_{A}}$ denote the indicator function of an event ${\displaystyle A}$, then ${\displaystyle \operatorname {E} [{\mathbf {1} }_{A}]=1\cdot \operatorname {P} (A)+0\cdot \operatorname {P} (\Omega \setminus A)=\operatorname {P} (A).}$
• Formulas in terms of CDF: If ${\displaystyle F(x)}$ is the cumulative distribution function of the probability measure ${\displaystyle \operatorname {P} ,}$ and ${\displaystyle X}$ is a random variable, then

${\displaystyle \operatorname {E} [X]=\int _{\overline {\mathbb {R} }}x\,dF(x),}$

where the values on both sides are well defined or not well defined simultaneously, and the integral is taken in the sense of Lebesgue-Stieltjes. Here, ${\displaystyle {\overline {\mathbb {R} }}=[-\infty ,+\infty ]}$ is the extended real line.

Additionally,

${\displaystyle \displaystyle \operatorname {E} [X]=\int \limits _{0}^{\infty }(1-F(x))\,dx-\int \limits _{-\infty }^{0}F(x)\,dx,}$

with the integrals taken in the sense of Lebesgue.

The proof of the second formula follows.

showProof.

For an arbitrary ${\displaystyle \omega \in \Omega ,}$ ${\displaystyle \displaystyle X^{-}(\omega )=\int \limits _{-X^{-}(\omega )}^{0}dx=\int \limits _{-\infty }^{0}{\mathbf {1} }{\{x\mid X^{-}(\omega )\geq -x\}}\,dx=\int \limits _{-\infty }^{0}{\mathbf {1} }{\{(\omega ,x)\mid X(\omega )\leq x\}}\,dx.}$ The last equality holds because the fact that ${\displaystyle X^{-}(\omega )\geq -x,}$ where ${\displaystyle x\leq 0,}$ implies that ${\displaystyle X^{+}(\omega )=0}$ and hence ${\displaystyle X(\omega )\leq x.}$ Conversely, if ${\displaystyle X(\omega )\leq x,}$ where ${\displaystyle x\leq 0,}$ then ${\displaystyle X^{+}(\omega )=0}$ and ${\displaystyle X^{-}(\omega )\geq -x.}$ The integrand in the above expression for ${\displaystyle X^{-}(\omega )}$ is non-negative, so Tonelli's theorem applies, and the order of integration may be switched without altering the result. We have ${\displaystyle {\begin{aligned}\operatorname {E} (X^{-})&=\int \limits _{\Omega }\left(\int \limits _{-\infty }^{0}{\mathbf {1} }{\{(\omega ,x)\mid X(\omega )\leq x\}}\,dx\right)d\operatorname {P} \\&=\int \limits _{-\infty }^{0}\left(\int \limits _{\Omega }{\mathbf {1} }{\{\omega \mid X(\omega )\leq x\}}\,d\operatorname {P} \right)dx\\&=\int \limits _{-\infty }^{0}\operatorname {P} (X\leq x)\,dx=\int \limits _{-\infty }^{0}F(x)\,dx.\end{aligned}}}$ Arguing as above, ${\displaystyle \displaystyle X^{+}(\omega )=\int \limits _{0}^{\infty }{\mathbf {1} }{\{(\omega ,x)\mid X(\omega )>x\}}\,dx,}$ and ${\displaystyle \displaystyle \operatorname {E} (X^{+})=\int \limits _{0}^{\infty }\operatorname {P} (X>x)\,dx=\int \limits _{0}^{\infty }(1-F(x))\,dx.}$ Recalling that ${\displaystyle \operatorname {E} (X)=\operatorname {E} (X^{+})-\operatorname {E} (X^{-})}$ completes the proof.

• Non-negativity: If ${\displaystyle X\geq 0}$ (a.s.), then ${\displaystyle \operatorname {E} [X]\geq 0}$.
• Linearity of expectation:^[5] The expected value operator (or expectation operator) ${\displaystyle \operatorname {E} [\cdot ]}$ is linear in the sense that, for any random variables ${\displaystyle X}$ and ${\displaystyle Y}$, and a constant ${\displaystyle a}$,

${\displaystyle {\begin{aligned}\operatorname {E} [X+Y]&=\operatorname {E} [X]+\operatorname {E} [Y],\\\operatorname {E} [aX]&=a\operatorname {E} [X],\end{aligned}}}$

whenever the right-hand side is well-defined. This means that the expected value of the sum of any finite number of random variables is the sum of the expected values of the individual random variables, and the expected value scales linearly with a multiplicative constant. Symbolically, for ${\displaystyle N}$ random variables ${\displaystyle X_{i}}$ and constants ${\displaystyle a_{i}(1\leq i\leq N)}$, we have ${\displaystyle \operatorname {E} \left[\sum _{i=1}^{N}a_{i}X_{i}\right]=\sum _{i=1}^{N}a_{i}\operatorname {E} [X_{i}]}$.

• Monotonicity: If ${\displaystyle X\leq Y}$ (a.s.), and both ${\displaystyle \operatorname {E} [X]}$ and ${\displaystyle \operatorname {E} [Y]}$ exist, then ${\displaystyle \operatorname {E} [X]\leq \operatorname {E} [Y]}$.

Proof follows from the linearity and the non-negativity property for ${\displaystyle Z=Y-X}$, since ${\displaystyle Z\geq 0}$ (a.s.).

• Non-multiplicativity: In general, the expected value is not multiplicative, i.e. ${\displaystyle \operatorname {E} [XY]}$ is not necessarily equal to ${\displaystyle \operatorname {E} [X]\cdot \operatorname {E} [Y]}$. If ${\displaystyle X}$ and ${\displaystyle Y}$ are independent, then one can show that ${\displaystyle \operatorname {E} [XY]=\operatorname {E} [X]\operatorname {E} [Y]}$. If the random variables are dependent, then generally ${\displaystyle \operatorname {E} [XY]\neq \operatorname {E} [X]\operatorname {E} [Y]}$, although in special cases of dependency the equality may hold.
• Law of the unconscious statistician: The expected value of a measurable function of ${\displaystyle X}$, ${\displaystyle g(X)}$, given that ${\displaystyle X}$ has a probability density function ${\displaystyle f(x)}$, is given by the inner product of ${\displaystyle f}$ and ${\displaystyle g}$:

${\displaystyle \operatorname {E} [g(X)]=\int _{\mathbb {R} }g(x)f(x)\,dx.}$ ^[5]

This formula also holds in multidimensional case, when ${\displaystyle g}$ is a function of several random variables, and ${\displaystyle f}$ is their joint density.^[5][15]

• Non-degeneracy: If ${\displaystyle \operatorname {E} [|X|]=0}$, then ${\displaystyle X=0}$ (a.s.).
• For a random variable ${\displaystyle X}$ with well-defined expectation: ${\displaystyle |\operatorname {E} [X]|\leq \operatorname {E} |X|}$.
• The following statements regarding a random variable ${\displaystyle X}$ are equivalent:
  • ${\displaystyle \operatorname {E} [X]}$ exists and is finite.
  • Both ${\displaystyle \operatorname {E} [X^{+}]}$ and ${\displaystyle \operatorname {E} [X^{-}]}$ are finite.
  • ${\displaystyle \operatorname {E} [|X|]}$ is finite.

For the reasons above, the expressions "${\displaystyle X}$ is integrable" and "the expected value of ${\displaystyle X}$ is finite" are used interchangeably throughout this article.

• If ${\displaystyle \operatorname {E} [X]<+\infty }$ then ${\displaystyle X<+\infty }$ (a.s.). Similarly, if ${\displaystyle \operatorname {E} [X]>-\infty }$ then ${\displaystyle X>-\infty }$ (a.s.).
• If ${\displaystyle \operatorname {E} |X^{\beta }|<\infty }$ and ${\displaystyle 0<\alpha <\beta ,}$ then ${\displaystyle \operatorname {E} |X^{\alpha }|<\infty .}$
• If ${\displaystyle X=Y}$ (a.s.), then ${\displaystyle \operatorname {E} [X]=\operatorname {E} [Y]}$. In other words, if X and Y are random variables that take different values with probability zero, then the expectation of X will equal the expectation of Y.
• If ${\displaystyle X=c}$ (a.s.) for some constant ${\displaystyle c\in [-\infty ,+\infty ]}$, then ${\displaystyle \operatorname {E} [X]=c}$. In particular, for a random variable ${\displaystyle X}$ with well-defined expectation, ${\displaystyle \operatorname {E} [\operatorname {E} [X]]=\operatorname {E} [X]}$. A well defined expectation implies that there is one number, or rather, one constant that defines the expected value. Thus follows that the expectation of this constant is just the original expected value.
• For a non-negative integer-valued random variable ${\displaystyle X:\Omega \to \{0,1,2,3,\ldots ,+\infty \},}$

${\displaystyle \operatorname {E} [X]=\sum _{n=0}^{\infty }\operatorname {P} (X>n).}$

showProof.

If ${\displaystyle \operatorname {P} (X=+\infty )>0,}$ then ${\displaystyle \operatorname {E} [X]=+\infty .}$ On the other hand, ${\displaystyle \operatorname {P} (X>n)\geq \operatorname {P} (X=+\infty )>0,}$ so the series on the right diverges to ${\displaystyle +\infty ,}$ and the equality holds. If ${\displaystyle \operatorname {P} (X=+\infty )=0,}$ then ${\displaystyle \sum _{n=0}^{\infty }\operatorname {P} (X>n)=\sum _{n=0}^{\infty }\sum _{j=n+1}^{\infty }\operatorname {P} (X=j).}$ Define the infinite upper-triangular matrix ${\displaystyle M={\begin{bmatrix}\operatorname {P} (X=1)&\operatorname {P} (X=2)&\operatorname {P} (X=3)&\cdots &\operatorname {P} (X=n)&\cdots \\&\operatorname {P} (X=2)&\operatorname {P} (X=3)&\cdots &\operatorname {P} (X=n)&\cdots \\&&\operatorname {P} (X=3)&\cdots &\operatorname {P} (X=n)&\cdots \\&&&\ddots &\vdots &\\&&&&\operatorname {P} (X=n)&\cdots \\&&&&&\ddots \end{bmatrix}}.}$ The double series ${\displaystyle \textstyle \sum _{i=1}^{\infty }\sum _{j=i}^{\infty }\operatorname {P} (X=j)}$ is the sum of ${\displaystyle M}$'s elements if summation is done row by row. Since every summand is non-negative, the series either converges absolutely or diverges to ${\displaystyle +\infty .}$ In both cases, changing summation order does not affect the sum. Changing summation order, from row-by-row to column-by-column, gives us ${\displaystyle {\begin{aligned}\sum _{n=0}^{\infty }\sum _{j=n+1}^{\infty }\operatorname {P} (X=j)&=\sum _{j=1}^{\infty }\sum _{n=0}^{j-1}\operatorname {P} (X=j)\\&=\sum _{j=1}^{\infty }j\operatorname {P} (X=j)\\&=\sum _{j=0}^{\infty }j\operatorname {P} (X=j)\\&=\operatorname {E} [X].\end{aligned}}}$

Uses and applications[]

The expectation of a random variable plays an important role in a variety of contexts. For example, in decision theory, an agent making an optimal choice in the context of incomplete information is often assumed to maximize the expected value of their utility function. For a different example, in statistics, where one seeks estimates for unknown parameters based on available data, the estimate itself is a random variable. In such settings, a desirable criterion for a "good" estimator is that it is unbiased; that is, the expected value of the estimate is equal to the true value of the underlying parameter.

It is possible to construct an expected value equal to the probability of an event, by taking the expectation of an indicator function that is one if the event has occurred and zero otherwise. This relationship can be used to translate properties of expected values into properties of probabilities, e.g. using the law of large numbers to justify estimating probabilities by frequencies.

The expected values of the powers of X are called the moments of X; the moments about the mean of X are expected values of powers of X − E[X]. The moments of some random variables can be used to specify their distributions, via their moment generating functions.

To empirically estimate the expected value of a random variable, one repeatedly measures observations of the variable and computes the arithmetic mean of the results. If the expected value exists, this procedure estimates the true expected value in an unbiased manner and has the property of minimizing the sum of the squares of the residuals (the sum of the squared differences between the observations and the estimate). The law of large numbers demonstrates (under fairly mild conditions) that, as the size of the sample gets larger, the variance of this estimate gets smaller.

This property is often exploited in a wide variety of applications, including general problems of statistical estimation and machine learning, to estimate (probabilistic) quantities of interest via Monte Carlo methods, since most quantities of interest can be written in terms of expectation, e.g. ${\displaystyle \operatorname {P} ({X\in {\mathcal {A}}})=\operatorname {E} [{\mathbf {1} }_{\mathcal {A}}]}$, where ${\displaystyle {\mathbf {1} }_{\mathcal {A}}}$ is the indicator function of the set ${\displaystyle {\mathcal {A}}}$.

The mass of probability distribution is balanced at the expected value, here a Beta(α,β) distribution with expected value α/(α+β).

In classical mechanics, the center of mass is an analogous concept to expectation. For example, suppose X is a discrete random variable with values x_i and corresponding probabilities p_i. Now consider a weightless rod on which are placed weights, at locations x_i along the rod and having masses p_i (whose sum is one). The point at which the rod balances is E[X].

Expected values can also be used to compute the variance, by means of the computational formula for the variance

${\displaystyle \operatorname {Var} (X)=\operatorname {E} [X^{2}]-(\operatorname {E} [X])^{2}.}$

A very important application of the expectation value is in the field of quantum mechanics. The expectation value of a quantum mechanical operator ${\displaystyle {\hat {A}}}$ operating on a quantum state vector ${\displaystyle |\psi \rangle }$ is written as ${\displaystyle \langle {\hat {A}}\rangle =\langle \psi |A|\psi \rangle }$. The uncertainty in ${\displaystyle {\hat {A}}}$ can be calculated using the formula ${\displaystyle (\Delta A)^{2}=\langle {\hat {A}}^{2}\rangle -\langle {\hat {A}}\rangle ^{2}}$.

Interchanging limits and expectation[]

In general, it is not the case that ${\displaystyle \operatorname {E} [X_{n}]\to \operatorname {E} [X]}$ despite ${\displaystyle X_{n}\to X}$ pointwise. Thus, one cannot interchange limits and expectation, without additional conditions on the random variables. To see this, let ${\displaystyle U}$ be a random variable distributed uniformly on ${\displaystyle [0,1]}$. For ${\displaystyle n\geq 1,}$ define a sequence of random variables

${\displaystyle X_{n}=n\cdot \mathbf {1} \left\{U\in \left[0,{\tfrac {1}{n}}\right]\right\},}$

with ${\displaystyle {\mathbf {1} }\{A\}}$ being the indicator function of the event ${\displaystyle A}$. Then, it follows that ${\displaystyle X_{n}\to 0}$ (a.s). But, ${\displaystyle \operatorname {E} [X_{n}]=n\cdot \operatorname {P} \left(U\in \left[0,{\tfrac {1}{n}}\right]\right)=n\cdot {\tfrac {1}{n}}=1}$ for each ${\displaystyle n}$. Hence, ${\displaystyle \lim _{n\to \infty }\operatorname {E} [X_{n}]=1\neq 0=\operatorname {E} \left[\lim _{n\to \infty }X_{n}\right].}$

Analogously, for general sequence of random variables ${\displaystyle \{Y_{n}:n\geq 0\}}$, the expected value operator is not ${\displaystyle \sigma }$-additive, i.e.

${\displaystyle \operatorname {E} \left[\sum _{n=0}^{\infty }Y_{n}\right]\neq \sum _{n=0}^{\infty }\operatorname {E} [Y_{n}].}$

An example is easily obtained by setting ${\displaystyle Y_{0}=X_{1}}$ and ${\displaystyle Y_{n}=X_{n+1}-X_{n}}$ for ${\displaystyle n\geq 1}$, where ${\displaystyle X_{n}}$ is as in the previous example.

A number of convergence results specify exact conditions which allow one to interchange limits and expectations, as specified below.

• Monotone convergence theorem: Let ${\displaystyle \{X_{n}:n\geq 0\}}$ be a sequence of random variables, with ${\displaystyle 0\leq X_{n}\leq X_{n+1}}$ (a.s) for each ${\displaystyle n\geq 0}$. Furthermore, let ${\displaystyle X_{n}\to X}$ pointwise. Then, the monotone convergence theorem states that ${\displaystyle \lim _{n}\operatorname {E} [X_{n}]=\operatorname {E} [X].}$

Using the monotone convergence theorem, one can show that expectation indeed satisfies countable additivity for non-negative random variables. In particular, let ${\displaystyle \{X_{i}\}_{i=0}^{\infty }}$ be non-negative random variables. It follows from monotone convergence theorem that

${\displaystyle \operatorname {E} \left[\sum _{i=0}^{\infty }X_{i}\right]=\sum _{i=0}^{\infty }\operatorname {E} [X_{i}].}$

• Fatou's lemma: Let ${\displaystyle \{X_{n}\geq 0:n\geq 0\}}$ be a sequence of non-negative random variables. Fatou's lemma states that

${\displaystyle \operatorname {E} [\liminf _{n}X_{n}]\leq \liminf _{n}\operatorname {E} [X_{n}].}$

Corollary. Let ${\displaystyle X_{n}\geq 0}$ with ${\displaystyle \operatorname {E} [X_{n}]\leq C}$ for all ${\displaystyle n\geq 0}$. If ${\displaystyle X_{n}\to X}$ (a.s), then ${\displaystyle \operatorname {E} [X]\leq C.}$

Proof is by observing that ${\displaystyle \textstyle X=\liminf _{n}X_{n}}$ (a.s.) and applying Fatou's lemma.

• Dominated convergence theorem: Let ${\displaystyle \{X_{n}:n\geq 0\}}$ be a sequence of random variables. If ${\displaystyle X_{n}\to X}$ pointwise (a.s.), ${\displaystyle |X_{n}|\leq Y\leq +\infty }$ (a.s.), and ${\displaystyle \operatorname {E} [Y]<\infty }$. Then, according to the dominated convergence theorem,
  • ${\displaystyle \operatorname {E} |X|\leq \operatorname {E} [Y]<\infty }$;
  • ${\displaystyle \lim _{n}\operatorname {E} [X_{n}]=\operatorname {E} [X]}$
  • ${\displaystyle \lim _{n}\operatorname {E} |X_{n}-X|=0.}$
• Uniform integrability: In some cases, the equality ${\displaystyle \displaystyle \lim _{n}\operatorname {E} [X_{n}]=\operatorname {E} [\lim _{n}X_{n}]}$ holds when the sequence ${\displaystyle \{X_{n}\}}$ is uniformly integrable.

Inequalities[]

There are a number of inequalities involving the expected values of functions of random variables. The following list includes some of the more basic ones.

• Markov's inequality: For a nonnegative random variable ${\displaystyle X}$ and ${\displaystyle a>0}$, Markov's inequality states that

${\displaystyle \operatorname {P} (X\geq a)\leq {\frac {\operatorname {E} [X]}{a}}.}$

• Bienaymé-Chebyshev inequality: Let ${\displaystyle X}$ be an arbitrary random variable with finite expected value ${\displaystyle \operatorname {E} [X]}$ and finite variance ${\displaystyle \operatorname {Var} [X]\neq 0}$. The Bienaymé-Chebyshev inequality states that, for any real number ${\displaystyle k>0}$,

${\displaystyle \operatorname {P} {\Bigl (}{\Bigl |}X-\operatorname {E} [X]{\Bigr |}\geq k{\sqrt {\operatorname {Var} [X]}}{\Bigr )}\leq {\frac {1}{k^{2}}}.}$

• Jensen's inequality: Let ${\displaystyle f:{\mathbb {R} }\to {\mathbb {R} }}$ be a Borel convex function and ${\displaystyle X}$ a random variable such that ${\displaystyle \operatorname {E} |X|<\infty }$. Then

${\displaystyle f(\operatorname {E} (X))\leq \operatorname {E} (f(X)).}$

The right-hand side is well defined even if ${\displaystyle X}$ assumes non-finite values. Indeed, as noted above, the finiteness of ${\displaystyle \operatorname {E} |X|}$ implies that ${\displaystyle X}$ is finite a.s.; thus ${\displaystyle f(X)}$ is defined a.s..

• Lyapunov's inequality:^[16] Let ${\displaystyle 0<s<t}$. Lyapunov's inequality states that

${\displaystyle \left(\operatorname {E} |X|^{s}\right)^{1/s}\leq \left(\operatorname {E} |X|^{t}\right)^{1/t}.}$

Proof. Applying Jensen's inequality to ${\displaystyle |X|^{s}}$ and ${\displaystyle g(x)=|x|^{t/s}}$, obtain ${\displaystyle {\Bigl |}\operatorname {E} |X^{s}|{\Bigr |}^{t/s}\leq \operatorname {E} |X^{s}|^{t/s}=\operatorname {E} |X|^{t}}$. Taking the ${\displaystyle t^{th}}$ root of each side completes the proof.

• Cauchy–Bunyakovsky–Schwarz inequality: The Cauchy–Bunyakovsky–Schwarz inequality states that

${\displaystyle (\operatorname {E} [XY])^{2}\leq \operatorname {E} [X^{2}]\cdot \operatorname {E} [Y^{2}].}$

• Hölder's inequality: Let ${\displaystyle p}$ and ${\displaystyle q}$ satisfy ${\displaystyle 1\leq p\leq \infty }$, ${\displaystyle 1\leq q\leq \infty }$, and ${\displaystyle 1/p+1/q=1}$. The Hölder's inequality states that

${\displaystyle \operatorname {E} |XY|\leq (\operatorname {E} |X|^{p})^{1/p}(\operatorname {E} |Y|^{q})^{1/q}.}$

• Minkowski inequality: Let ${\displaystyle p}$ be a positive real number satisfying ${\displaystyle 1\leq p\leq \infty }$. Let, in addition, ${\displaystyle \operatorname {E} |X|^{p}<\infty }$ and ${\displaystyle \operatorname {E} |Y|^{p}<\infty }$. Then, according to the Minkowski inequality, ${\displaystyle \operatorname {E} |X+Y|^{p}<\infty }$ and

${\displaystyle {\Bigl (}\operatorname {E} |X+Y|^{p}{\Bigr )}^{1/p}\leq {\Bigl (}\operatorname {E} |X|^{p}{\Bigr )}^{1/p}+{\Bigl (}\operatorname {E} |Y|^{p}{\Bigr )}^{1/p}.}$

Expected values of common distributions[]

Distribution	Notation	Mean E(X)
Bernoulli	${\displaystyle X\sim ~b(1,p)}$	${\displaystyle p}$
Binomial	${\displaystyle X\sim B(n,p)}$	${\displaystyle np}$
Poisson	${\displaystyle X\sim Po(\lambda )}$	${\displaystyle \lambda }$
Geometric	${\displaystyle X\sim Geometric(p)}$	${\displaystyle 1/p}$
Uniform	${\displaystyle X\sim U(a,b)}$	${\displaystyle (a+b)/2}$
Exponential	${\displaystyle X\sim \exp(\lambda )}$	${\displaystyle 1/\lambda }$
Normal	${\displaystyle X\sim N(\mu ,\sigma ^{2})}$	${\displaystyle \mu }$
Standard Normal	${\displaystyle X\sim N(0,1)}$	${\displaystyle 0}$
Pareto	${\displaystyle X\sim Par(\alpha )}$	${\displaystyle \alpha /(\alpha +1)}$ if ${\displaystyle \alpha >1}$
Cauchy	${\displaystyle X\sim Cauchy(x_{0},\gamma )}$	undefined

Relationship with characteristic function[]

The probability density function ${\displaystyle f_{X}}$ of a scalar random variable ${\displaystyle X}$ is related to its characteristic function ${\displaystyle \varphi _{X}}$ by the inversion formula:

${\displaystyle f_{X}(x)={\frac {1}{2\pi }}\int _{\mathbb {R} }e^{-itx}\varphi _{X}(t)\,\mathrm {d} t.}$

For the expected value of ${\displaystyle g(X)}$ (where ${\displaystyle g:{\mathbb {R} }\to {\mathbb {R} }}$ is a Borel function), we can use this inversion formula to obtain

${\displaystyle \operatorname {E} [g(X)]={\frac {1}{2\pi }}\int _{\mathbb {R} }g(x)\left[\int _{\mathbb {R} }e^{-itx}\varphi _{X}(t)\,\mathrm {d} t\right]\,\mathrm {d} x.}$

If ${\displaystyle \operatorname {E} [g(X)]}$ is finite, changing the order of integration, we get, in accordance with Fubini–Tonelli theorem,

${\displaystyle \operatorname {E} [g(X)]={\frac {1}{2\pi }}\int _{\mathbb {R} }G(t)\varphi _{X}(t)\,\mathrm {d} t,}$

where

${\displaystyle G(t)=\int _{\mathbb {R} }g(x)e^{-itx}\,\mathrm {d} x}$

is the Fourier transform of ${\displaystyle g(x).}$ The expression for ${\displaystyle \operatorname {E} [g(X)]}$ also follows directly from Plancherel theorem.

See also[]

• Center of mass
• Central tendency
• Chebyshev's inequality (an inequality on location and scale parameters)
• Conditional expectation
• Expectation (the general term)
• Expectation value (quantum mechanics)
• Law of total expectation—the expected value of the conditional expected value of X given Y is the same as the expected value of X.
• Moment (mathematics)
• Nonlinear expectation (a generalization of the expected value)
• Wald's equation—an equation for calculating the expected value of a random number of random variables

References[]

Literature[]