Expected value is a fundamental concept in the category of probability in mathematics. The expected value of a variable is defined as the average value which is intuitive in long run value by repeatedly performing the experiment which it is representing. Other names for expected value are mathematical expectation, mean, expectation, first moment or EV.

We denote the expected value of a function f (x) by < f (x)> or E [f (x)]
In other words we can say that the expectation of a discrete kind of random variable is simply the weighted average probability of the values that are possible. 

The expected value plays an important role in characterization of the probability distribution. It serves as a key aspect since it is one of the location parameter type. 


Few properties of expected value are as:
1) The expected value of some constant is same as the value of the constant. This implies that E [b] = b, where ‘b is a constant.

2) If M <= N are random variables given, then it implies that E [M] <= E [N].

3) The operator of expectation is linear.

E [M + b] = E [M] + b

E [M + N] = E [M] + E [N]

E [b M] = b E [M]

Where ‘b’ is a constant and M and N are random variables.
For a discrete and single variable we define is as follows:

<f (x)> = $\sum_{x}$ f (x) P (x)

Where, P (x) represents the probability density function.

For a continuous and single variable we define it as follows:
<f (x)> = $\int$ f (x) P (x) dx
For discrete and multiple variables we follow

$<f (x_1, x_2,…., x_n)>$ = $\sum_{x_1, x_2,…, x_n} f (x_1, …, x_n) P (x_1, x_2,…, x_n)$

For continuous and multiple variables we follow

$<f (x_1, x_2,…., x_n)>$ = $\int$ $f (x_1, …, x_n) P (x_1, x_2,…, x_n) dx_1 dx_2 …. dx_n$


Below we have discussed three methods to find the expected value:

The first method follows following steps:
1) First we get ourselves familiar with the given problem.
2) Then we list out all the possible results of that problem.
3) Thereafter we calculate the probability of every outcome.
4) The value of each result is then written down.
5) The event’s expected value is then calculated by adding the value of every result together.
6) Understanding of the implications of the calculation of expected value.

The second method follows given steps:

1) Find the average number of times the experiment to be repeated to get the desired pattern.
2) Assume the required answer to be ‘x’.
3) Consider the possibilities of first trial and what will happen on each trial.
4) Find the results with different combinations of simultaneous occurrence of events.
5) Simplify the equation and solve for x.

The third method uses the given steps:

1) One should know what is an expected value.
2) Independent events should also be known.
3) The law of large numbers must be clear.
4) By the understanding of above three steps one can judge on the expected value.
Any of the above methods can be used to find the expected value of a distribution. Using expected value we can also find the variance and standard deviation of any function.


Below are some example s on expected value:

Example 1: Find the expected value for the random variable of an unbiased die.

Solution: The sample space for a die = {1, 2, 3, 4, 5, 6}

Let x be a random variable, i.e. up face of a die, so the probability mass function, say g(x) = $\frac{1}{6}$

Now the expected value of x is

E(x) = 1 x $\frac{1}{6}$ + 2 x $\frac{1}{6}$ + 3 x $\frac{1}{6}$ + 4 x $\frac{1}{6}$ + 5 x $\frac{1}{6}$ + 6 x $\frac{1}{6}$

= $\frac{1}{6}$ + $\frac{2}{6}$ + $\frac{3}{6}$ + $\frac{4}{6}$ + $\frac{5}{6}$ + $\frac{6}{6}$

= $\frac{21}{6}$

Example 2: The following table is describing about the probability mass function for random variable x.

 g(x)  0.1  0.1  0.2

Find the standard deviation of function.


g(x) 0.1 0.1 0.2
 x-$\mu$ 1.3 2.3 3.3
 (x-$\mu$)$^2$ 1.69 5.29 10.89

 First find expected value from the given data:

E(x) = 3 x 0.1 + 4 x 0.1 + 5 x 0.2 = 1.7

or $\mu$ = 1.7

Now $\sigma^2$ = E(x - $\mu$)$^2$

= 1.69x 0.1 + 5.29 x 0.1 + 10.89 x 0.2

= 0.169 + 0.529 + 2.178

= 2.876

Standard deviation = $\sqrt{ 2.876}$ = 1.70 (approx)