Variance, Intuitively

The Expected Value tells you where a random outcome sits on average. Variance tells you how much it moves around that average. Two bets can share the exact same expected value and still feel completely different: one barely wobbles, the other swings wildly. Variance is the number that separates them, and it's the starting point for every honest conversation about risk.

The idea in words: for each possible outcome, measure how far it lands from the mean, square that distance, and take the average of those squared distances. Squaring does two jobs at once, it makes every distance positive (so overs and unders don't cancel), and it punishes big misses far more than small ones.

If a random quantity $X$ has mean $\mu = E[X]$, its variance is

$$\mathrm{Var}(X) = E\big[(X - \mu)^2\big] = \sum_i p_i\,(x_i - \mu)^2,$$

where $x_i$ are the possible values and $p_i$ their probabilities. Every symbol here you already know from expected value: this is just the expected value of the squared distance from the mean.

Notice the units. If $X$ is measured in dollars, variance comes out in dollars squared, an awkward unit that doesn't compare directly to the outcomes. That's exactly why Standard Deviation exists: take the square root of the variance and you're back in dollars, in a number you can lay next to the mean.

Worked example: same mean, very different risk

Bet A pays $+1$ or $-1$, each with probability $\tfrac12$. Bet B pays $+10$ or $-10$, each with probability $\tfrac12$. Both have mean $\mu = 0$.

For Bet A, the squared distances from the mean are $(1-0)^2 = 1$ and $(-1-0)^2 = 1$, so

$$\mathrm{Var}(A) = \tfrac12(1) + \tfrac12(1) = 1 .$$

For Bet B, the squared distances are $100$ and $100$, so

$$\mathrm{Var}(B) = \tfrac12(100) + \tfrac12(100) = 100 .$$

Same expected value, but Bet B's variance is $100$ times larger. Its standard deviation ($\sqrt{100}=10$) is ten times Bet A's ($\sqrt{1}=1$), which matches the intuition that its swings are ten times bigger.

A handy shortcut

Computing $(x_i-\mu)^2$ for every outcome is tedious. The algebra collapses to a formula you'll use constantly:

$$\mathrm{Var}(X) = E[X^2] - \big(E[X]\big)^2 .$$

Take a fair die: $E[X] = 3.5$, and $E[X^2] = \tfrac{1+4+9+16+25+36}{6} = \tfrac{91}{6} \approx 15.17$. So $\mathrm{Var}(X) = 15.17 - 3.5^2 = 15.17 - 12.25 \approx 2.92$. The average squared distance of a die roll from $3.5$ is just under $3$.

Common pitfalls

• Variance is not in the same units as the data. It's squared. Don't compare a variance directly to a mean or a price, convert to Standard Deviation first.
• Same mean does not mean same risk. The whole point of the diagram above: two bets with identical expected value can carry wildly different variance. Never size a position off the average alone.
• The shortcut needs $E[X^2]$, not $E[X]^2$ done first. A classic slip is computing $E[X]$, squaring it, and forgetting that $E[X^2]$ is a separate expectation over the squared values.
• Outliers dominate. Because distances are squared, one rare huge outcome can swell the variance far more than many small ones. That sensitivity is a feature for risk, but it means fat tails inflate variance fast.

Once you can compute a mean and a variance for any bet, you have the two numbers that anchor almost every quant answer. The variance feeds directly into Standard Deviation, into how outcomes move together, and into the higher moments that describe the finer shape of risk.