Expected value

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In probability theory the expected value (or mathematical expectation, or mean) of a discrete random variable is the sum of the probability of each possible outcome of the experiment multiplied by the outcome value (or payoff). Thus, it represents the average amount one "expects" as the outcome of the random trial when identical odds are repeated many times. Note that the value itself may not be expected in the general sense - the "expected value" itself may be unlikely or even impossible.

For example, the expected value from the roll of an ordinary six-sided die is 3.5, found by,

$\begin{align} \operatorname{E}(X)& = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6}\\[6pt] & = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5, \end{align}$

which is not one of the possible outcomes.

A common application of expected value is in gambling. For example, an American roulette wheel has 38 equally likely outcomes. A winning bet placed on a single number pays 35-to-1 (this means that you are paid 35 times your bet and your bet is returned, so you get 36 times your bet). So considering all 38 possible outcomes, the expected value of the profit resulting from a $1 bet on a single number is:

$\left( -\$1 \times \frac{37}{38} \right) + \left( \$35 \times \frac{1}{38} \right),$

which is about −$0.0526. (Your net is -$1 when you lose and $35 when you win.) Therefore one expects, on average, to lose over five cents for every dollar bet, and the expected value of a one dollar bet is $0.9473. In gambling or betting, a game or situation in which the expected value of the profit for the player is zero (no net gain nor loss) is commonly called a "fair game."

1 Mathematical definition
2 Conventional terminology
3 Properties
4 Uses and applications of the expected value
5 Expectation of matrices
6 Computation
7 Formula for Non-Negative Integral Values
8 See also
9 External links

In general, if $X\,$ is a random variable defined on a probability space $(\Omega, \Sigma, P)\,$ , then the expected value of $X\,$ (denoted $\operatorname{E}(X)\,$ or sometimes $\langle X \rangle$ or $\mathbb{E}(X)$ ) is defined as

$\operatorname{E}(X) = \int_\Omega X\, \operatorname{d}P\,$

where the Lebesgue integral is employed. Note that not all random variables have an expected value, since the integral may not exist (e.g., Cauchy distribution). Two variables with the same probability distribution will have the same expected value, if it is defined.

$\operatorname{E}(X) = \sum_i p_i x_i\,$

as in the gambling example mentioned above.

If the probability distribution of $X$ admits a probability density function $f (x)$ , then the expected value can be computed as

$\operatorname{E}(X) = \int_{-\infty}^\infty x f(x)\, \operatorname{d}x .$

It follows directly from the discrete case definition that if $X$ is a constant random variable, i.e. $X = b$ for some fixed real number $b$ , then the expected value of $X$ is also $b$ .

The expected value of an arbitrary function of X, g(X), with respect to the probability density function f(x) is given by:

$\operatorname{E}(g(X)) = \int_{-\infty}^\infty g(x) f(x)\, \operatorname{d}x .$

When one speaks of the "expected price", "expected height", etc. one means the expected value of a random variable that is a price, a height, etc.
When one speaks of the "expected number of attempts needed to get one successful attempt," one might conservatively approximate it as the reciprocal of the probability of success for such an attempt.^{[citation needed]}

The expected value of a constant is equal to the constant itself; i.e., if c is a constant, then E(c) = c

If X and Y are random variables so that $X \le Y$ almost surely, then $\operatorname{E}(X) \le \operatorname{E}(Y)$ .

The expected value operator (or expectation operator) $\operatorname{E}$ is linear in the sense that

$\operatorname{E}(X + c)= \operatorname{E}(X) + c\,$

$\operatorname{E}(X + Y)= \operatorname{E}(X) + \operatorname{E}(Y)\,$

$\operatorname{E}(aX)= a \operatorname{E}(X)\,$

Combining the results from previous three equations, we can see that -

$\operatorname{E}(aX + b)= a \operatorname{E}(X) + b\,$

$\operatorname{E}(a X + b Y) = a \operatorname{E}(X) + b \operatorname{E}(Y)\,$

for any two random variables $X$ and $Y$ (which need to be defined on the same probability space) and any real numbers $a$ and $b$ .

For any two discrete random variables $X, Y$ one may define the conditional expectation:

$\operatorname{E}(X|Y)(y) = \operatorname{E}(X|Y=y) = \sum\limits_x x \cdot \operatorname{P}(X=x|Y=y).$

which means that $\operatorname{E}(X|Y)$ is a function on $y$ .

Then the expectation of $X$ satisfies

$\operatorname{E} \left( \operatorname{E}(X|Y) \right)= \sum\limits_y \operatorname{E}(X|Y=y) \cdot \operatorname{P}(Y=y) \,$

$=\sum\limits_y \left( \sum\limits_x x \cdot \operatorname{P}(X=x|Y=y) \right) \cdot \operatorname{P}(Y=y)\,$

$=\sum\limits_y \sum\limits_x x \cdot \operatorname{P}(X=x|Y=y) \cdot \operatorname{P}(Y=y)\,$

$=\sum\limits_y \sum\limits_x x \cdot \operatorname{P}(Y=y|X=x) \cdot \operatorname{P}(X=x) \,$

$=\sum\limits_x x \cdot \operatorname{P}(X=x) \cdot \left( \sum\limits_y \operatorname{P}(Y=y|X=x) \right) \,$

$=\sum\limits_x x \cdot \operatorname{P}(X=x) \,$

$=\operatorname{E}(X).\,$

Hence, the following equation holds:

$\operatorname{E}(X) = \operatorname{E} \left( \operatorname{E}(X|Y) \right).$

The right hand side of this equation is referred to as the iterated expectation and is also sometimes called the tower rule. This proposition is treated in law of total expectation.

In the continuous case, the results are completely analogous. The definition of conditional expectation would use inequalities, density functions, and integrals to replace equalities, mass functions, and summations, respectively. However, the main result still holds:

$\operatorname{E}(X) = \operatorname{E} \left( \operatorname{E}(X|Y) \right).$

If a random variable X is always less than or equal to another random variable Y, the expectation of X is less than or equal to that of Y:

If $X \leq Y$ , then $\operatorname{E}(X) \leq \operatorname{E}(Y)$ .

In particular, since $X \leq |X|$ and $-X \leq |X|$ , the absolute value of expectation of a random variable is less or equal to the expectation of its absolute value:

$|\operatorname{E}(X)| \leq \operatorname{E}(|X|)$

The following formula holds for any nonnegative real-valued random variable $X$ (such that $\operatorname{E}(X) < \infty$ ), and positive real number $α$ :

$\operatorname{E}(X^\alpha) = \alpha \int_{0}^{\infty} t^{\alpha -1}\operatorname{P}(X>t) \, \operatorname{d}t.$

In general, the expected value operator is not multiplicative, i.e. $\operatorname{E}(X Y)$ is not necessarily equal to $\operatorname{E}(X) \operatorname{E}(Y)$ . If multiplicativity occurs, the $X$ and $Y$ variables are said to be uncorrelated (independent variables are a notable case of uncorrelated variables). The lack of multiplicativity gives rise to study of covariance and correlation.

In general, the expectation operator and functions of random variables do not commute; that is

$\operatorname{E}(g(X)) = \int_{\Omega} g(X)\, \operatorname{d}P \neq g(\operatorname{E}(X)),$

A notable inequality concerning this topic is Jensen's inequality, involving expected values of convex (or concave) functions.

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 $X - \operatorname{E}(X)$ . 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. This 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.

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 $\operatorname{E}(X)$ .

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

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

If $X$ is an $m \times n$ matrix, then the expected value of the matrix is defined as the matrix of expected values:

$\operatorname{E}(X) = \operatorname{E} \begin{pmatrix} x_{1,1} & x_{1,2} & \cdots & x_{1,n} \\ x_{2,1} & x_{2,2} & \cdots & x_{2,n} \\ \vdots \\ x_{m,1} & x_{m,2} & \cdots & x_{m,n} \end{pmatrix} = \begin{pmatrix} \operatorname{E}(x_{1,1}) & \operatorname{E}(x_{1,2}) & \cdots & \operatorname{E}(x_{1,n}) \\ \operatorname{E}(x_{2,1}) & \operatorname{E}(x_{2,2}) & \cdots & \operatorname{E}(x_{2,n}) \\ \vdots \\ \operatorname{E}(x_{m,1}) & \operatorname{E}(x_{m,2}) & \cdots & \operatorname{E}(x_{m,n}) \end{pmatrix}.$

This is utilized in covariance matrices.

It is often useful to update a computed expected value as new data comes in. This can be done as follows, where $n e w_v a l u e$ is the $c o u n t$ -th value, and we use the previous estimate $\operatorname{E}_\mathrm{prev}(X)$ to compute $\operatorname{E}_\mathrm{new}(X)$ :

$\operatorname{E}_\mathrm{new}(X) = [(count-1) \cdot \operatorname{E}_\mathrm{prev}(X) + new\_value]/count$

When a random variable takes only values in ${0,1,2,3,...}$ we can use the following formula for computing its expectation:

$\operatorname{E}(X)=\sum\limits_{i=1}^\infty P(X\geq i)$

For example, suppose we toss a coin where the probability of heads is $p$ . How many tosses can we expect until the first heads? Let $X$ be this number. Note that we are counting only the tails and not the heads which ends the experiment; in particular, we can have $X = 0$ . The expectation of $X$ may be computed by $\sum_{i= 1}^\infty (1-p)^i=(1-p)/p$ . This is because the number of tosses is at least $i$ exactly when the first $i$ tosses yielded tails. This matches the expectation of a random variable with an Exponential distribution. We used the formula for Geometric progression: $\sum_{k=1}^\infty r^k=r/(1-r)$ .