# Why doesn't the Variance of a Random Variable use $\frac{1}{n}$ instead of $P(X=x_i)$? Aren't the weights already expressed in the Mean?

I understand that the population variance of a set is just a special case of the variance of a random variable where $P(X=x) = \frac{1}{n}$ for $n$ elements in the set.

Still I can't help but feel that when computing $\sigma^2 (X) = \sum_{i=1}^n [(x_i -\mu)^2 \cdot P(X=x_i)]$, we're doing extra, or somehow "too much", individual weighing of the values of $X$.

After all, aren't the weights by likelihood of occurrence of a certain $x_i$ already expressed in $\mu$? Why are we again weighing the squared differences from the mean by probability? Why not average using $\frac{1}{n}$ instead?

• Because then you'd be adding them instead of averaging them. The variance (loosely speaking) captures how much a variable may vary from its mean on average. – arctic tern Jul 3 '16 at 20:02
• @arctictern, In the title of the question I ask about using $\frac{1}{n}$ instead. So I'm not saying don't average them the differences, I'm asking why average using the individual probabilities, since they're already expressed in $\mu$, and why not use $\frac{1}{n}$ instead? – jeremy radcliff Jul 3 '16 at 20:03
• Why would we use $\frac{1}{n}$ if the different variations from the mean are not equally likely? – arctic tern Jul 3 '16 at 20:04
• @arctictern I don't know, I just feel that the likelihood unevenness is already captured in the mean, so it feels like overkill somehow to express that unevenness yet again. – jeremy radcliff Jul 3 '16 at 20:05
• Quote: "I'm asking why average using the individual probabilities ..." You don´t use the individual probability, but you divide by n. It is a coincindence that the individual probabilities are equal to $\frac1n$. – callculus Jul 3 '16 at 20:10

The squared variation of $X$ from its mean could itself be considered a random variable, say defined by $Y=(X-\mu)^2$. The variation then measures the mean of $Y$, which involves using $Y$'s and hence $X$'s probability distribution. Your feelings wouldn't say to average a random variable (like $Y$) by simply adding the possible values it could take and dividing by the count if $Y$ is not uniformly distributed would they? If $x_1$ and $x_2$ are two values that $X$ can take with some probability, then the squared variation $(X-\mu)^2$ is not equally likely to be $(x_1-\mu)^2$ as it is to be $(x_2-\mu)^2$ unless for instance $X$ itself is equally likely to be $x_1$ as it is to be $x_2$.

Consider for instance a random variable $X$ that took the values $a$ or $b$ with probabilities $1$ and $0$ respectively. So essentially, $X$ always takes the value $a$, and in particular its mean is $\mu=a$.

According to your feelings, we shouldn't take into account how likely it is that $X$ is $a$ versus $b$ and simply calculate the average squared variation to be

$$\frac{1}{2}(a-a)^2+\frac{1}{2}(b-a)^2=\frac{(b-a)^2}{2}.$$

In other words, the average squared variation of $X$ from its mean would be $(b-a)^2/2$, even though $X$ never varies from its mean value $\mu=a$ and the value $b$ is completely independent of $X$ anyway, which is absurd. The issue is that the squared variation $(X-\mu)^2$ is equal to $(a-a)^2$ one-hundred percent of the time (not $\frac{1}{2}$ of the time) and is equal to $(b-a)^2$ zero percent of the time (not $\frac{1}{2}$ of the time), so on average $(X-a)^2$ should be

$$1(a-a)^2+0(b-a)^2=0.$$

• Thank you for the added details. It took me a while to work through everything conceptually but it finally makes sense. – jeremy radcliff Jul 3 '16 at 22:35