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How to find the expected value M(X) and the variance D(X) from a distribution table

The rule: M(X) = Σ x·p, then D(X) = M(X²) − M(X)²

A discrete random variable is given by a table of its values and their probabilities. Three facts do all the work. The probabilities add up to $1$ - that recovers any missing entry. The expected value is the weighted average of the values:

M (X) = \sum x_{i} p_{i}

The variance measures how far the values scatter around that average. The definition is $M ((X - M (X))^{2})$ , but the short form is far quicker to compute:

D (X) = M (X^{2}) - (M (X))^{2}

where $M (X^{2})$ is found from the same table with each value squared and the probabilities untouched. Two sanity checks: $M (X)$ must lie between the smallest and the largest value, and $D (X)$ can never be negative.

Problem

A random variable $X$ is given by the distribution table

X P - 2 ? 0 0.5 3 0.3

Find $M (X)$ and $D (X)$ .

Answer

p = 0.2

M (X) = 0.5

D (X) = 3.25

(standard deviation

σ = 3.25 \approx 1.80

Checked by hand: the missing probability, the mean and the variance were all recomputed.

Step-by-step solution

1. Recover the missing probability

The probabilities of the whole table sum to $1$ :

p + 0.5 + 0.3 = 1

p = 1 - 0.8 = 0.2

So $P (X = - 2) = 0.2$ .

2. The expected value

Multiply each value by its probability and add:

M (X) = (- 2) \cdot 0.2 + 0 \cdot 0.5 + 3 \cdot 0.3

M (X) = - 0.4 + 0 + 0.9 = 0.5

The result sits between $- 2$ and $3$ , as an average must.

3. The variance through M(X²)

Square the values, keep the probabilities:

M (X^{2}) = (- 2)^{2} \cdot 0.2 + 0^{2} \cdot 0.5 + 3^{2} \cdot 0.3

M (X^{2}) = 0.8 + 0 + 2.7 = 3.5

Now the short formula:

D (X) = M (X^{2}) - (M (X))^{2} = 3.5 - (0.5)^{2}

D (X) = 3.5 - 0.25 = 3.25

4. If the standard deviation is asked too

σ (X) = D (X) = 3.25 \approx 1.80

Unlike the variance, it is measured in the same units as $X$ itself, which is why it is the number usually quoted.

How to check yourself

The long definition must give the same answer:

D (X) = (- 2 - 0.5)^{2} \cdot 0.2 + (0 - 0.5)^{2} \cdot 0.5 + (3 - 0.5)^{2} \cdot 0.3

= 6.25 \cdot 0.2 + 0.25 \cdot 0.5 + 6.25 \cdot 0.3 = 1.25 + 0.125 + 1.875 = 3.25

Same value - and if your two routes disagree, the error is almost always a forgotten square in $M (X^{2})$ .

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