Expectation, Variance & SD

A whole PMF table is more than anyone wants to carry around. Most of the time you only need two numbers: roughly where the variable sits, and roughly how much it swings. Those are the expectation (the average value if you repeated the experiment forever) and the variance (the average squared distance from that average). Take the square root of variance and you’re back in the original units — the standard deviation.

Pretty much every later lesson — binomial, Poisson, model error, the central limit theorem — is “what’s E[X] and Var(X) for this distribution?” GATE turns that into a NAT almost every year.

The three formulas

For a discrete random variable X with PMF p(x):

Two algebra rules turn these into quick answers without rebuilding sums:

• Linearity of expectation — E[aX + b] = aE[X] + b. Scaling and shifting the variable scales and shifts its mean, exactly. This holds always, even when variables are dependent.
• Variance under shift and scale — Var(aX + b) = a²·Var(X). A constant shift b moves every value equally, so it changes the mean but not the spread — b vanishes. A scale a stretches deviations, and since variance is squared distance it picks up a².

The variance formula itself is worth saying in words: the mean of the squares minus the square of the mean, Var(X) = E[X²] − (E[X])². The two pieces are different computations — E[X²] weights x² by p(x); (E[X])² squares the single number E[X].

A handy fact for the geometric setting: if each independent trial succeeds with probability p, the expected number of trials until the first success is 1/p. For a fair coin, p = 1/2, so you expect 1/(1/2) = 2 tosses to see the first head.

How GATE asks this

Overwhelmingly a NAT: a small PMF (or a fair die / coin) is given and you compute E[X], Var(X), or SD to a few decimals. The reliable route is a two-row table — one row for x·p(x) summing to E[X], one for x²·p(x) summing to E[X²] — then Var = E[X²] − (E[X])². The occasional MCQ tests the identities instead: which of E[aX+b] = aE[X]+b, Var(X+c) = Var(X), E[X²] = (E[X])² are always true.

Worked example — a fair six-sided die

A fair die shows 1–6, each with probability 1/6. Find E[X], E[X²], Var(X), and SD.

E[X]  = (1+2+3+4+5+6)/6        = 21/6  = 3.5

E[X²] = (1+4+9+16+25+36)/6     = 91/6  ≈ 15.1667

Var(X) = E[X²] − (E[X])²
       = 91/6 − 3.5²
       = 15.1667 − 12.25
       = 2.9167          (exactly 35/12)

SD = √2.9167 ≈ 1.708

Note the order: square each face and average for E[X²] = 15.1667, then subtract the square of the mean 3.5² = 12.25. Subtracting first or squaring the wrong quantity is the usual slip. As a second mini-example, the expected number of fair-coin tosses to get the first head is 1/(1/2) = 2.

Quick check