Calculate the expected value and variance of a discrete random variable

Primary Tool

Expected Value Calculator

The expected value calculator takes the full probability table and reports $E [X]$ , $Var (X)$ , and $σ$ in one step. That makes it easy to check each formula and to try variations by changing a single weight.

Problem

Let $X$ be the number of heads in three fair coin flips. Compute $E [X]$ , $Var (X)$ , and $σ_{X}$ .

Three coin flips produce a binomial distribution with $n = 3$ and $p = 0.5$ . Every outcome has a known probability from the binomial formula, which makes this a clean worked example for the general definitions.

The distribution table

List each outcome and its probability before computing anything.

• $X = 0$ : all tails. $P (X = 0) = (0.5)^{3} = 0.125$
• $X = 1$ : one head, two tails. $P (X = 1) = 3 (0.5)^{3} = 0.375$
• $X = 2$ : two heads, one tail. $P (X = 2) = 3 (0.5)^{3} = 0.375$
• $X = 3$ : all heads. $P (X = 3) = (0.5)^{3} = 0.125$

Step-by-step walkthrough

Work through expected value first, then use a second weighted sum to reach variance.

1Compute $E [X] = \sum x_{i} p_{i} = 0 (0.125) + 1 (0.375) + 2 (0.375) + 3 (0.125) = 1.5$ .
2Compute $E [X^{2}] = \sum x_{i}^{2} p_{i} = 0 (0.125) + 1 (0.375) + 4 (0.375) + 9 (0.125) = 3.0$ .
3Apply the shortcut formula: $Var (X) = E [X^{2}] - (E [X])^{2} = 3.0 - 1. 5^{2} = 3.0 - 2.25 = 0.75$ .
4Take the square root to get the standard deviation: $σ_{X} = 0.75 \approx 0.866$ .
5Confirm with the binomial formula: $n p = 3 (0.5) = 1.5$ and $n p (1 - p) = 3 (0.5) (0.5) = 0.75$ . Both match.

What variance is measuring

Variance $0.75$ says the outcomes are, on average, $0.75$ squared units away from the mean. Standard deviation $0.866$ puts that in the original units: a typical outcome lands less than one head away from the mean of $1.5$ .

For a three-flip experiment that matches intuition: most outcomes cluster around one or two heads, and exact zero or three heads each occur only $12.5%$ of the time.

Common Pitfall

The formula $Var (X) = E [X^{2}] - (E [X])^{2}$ is a computational shortcut for $\sum p_{i} (x_{i} - μ)^{2}$ . Both definitions give the same number, but the shortcut avoids computing each squared deviation by hand.

Try a Variation

Change the coin to one that lands heads with probability $p = 0.3$ instead of $0.5$ . Update the probability table and recompute $E [X]$ and $Var (X)$ . Do the results still match $n p$ and $n p (1 - p)$ ?

Keep moving through the cluster

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