Is the var$(y_i) = E(y_i^2)$?

Can someone tell me if it's true? Is it always? How so? Assuming that Y is a random variable with normal distribution with $\mu = \beta E(x_i)$ and $\sigma^2$.

Does it apply only to this case? Considering this that Y comes from $Y = \beta X + e$, where $e \sim N(0,\sigma^2$).


2 Answers 2


The general definition is:

$$ \operatorname{Var}(X) = \operatorname{E} \left[ (X - \operatorname{E}[X])^2 \right] $$

which can also be written as:

$$ \operatorname{E}\left[X^2 \right] - \operatorname{E}[X]^2 $$

In the special case where the mean is 0 (i.e. $E[X]=0$), you get the result that you provided.

More details here: https://en.wikipedia.org/wiki/Variance

I hope this helps.


By definition, for a random variable $X$, $Var(X) = E[(X-E[X])^2] = \sum_{x} (X - E[X])^2 P(X=x)$.

We can write this formula in what is often a more convenient way: $Var(X) = E[X^2] - E^2[X]$.

The derivation of this follows below:

$$Var(X) = \sum_{x} (X - E[X])^2 P(X=x) = \sum_{x} (X^2 - 2XE[X] + E^2[X]) P(X=x) = \sum_{x} X^2 P(X=x) - 2E[X] \sum_{x} X P(X=x) + E^2[X] = E[X^2] - 2E^2[X] + E^2[X] = E[X^2] - E^2[X]$$

In the case of a normal distribution of mean (expectation) $\mu = \beta x_i$ and variance $\sigma^2$ (where $\sigma$ is the standard deviation), the expression you mentioned is not true.

To see this, let us attempt a proof by contradiction.

Let's assume that the statement is true: $Var(y_i) = E(y_i^2)$.

Then, we have that $E^2[y_i] = 0$, which implies that $E[y_i] = 0$.

However, we know that $E[y_i] = \beta E[x_i]$. So, unless $\beta$ or $E[x_i]$ are equal to zero, it doesn't hold, since we have a contradiction. The formula for the expectation of $y_i$ can be derived from the formula for $Y$ and the linearity of expectation.

The only case when what you state is true is when the expectation of the random variable is indeed $0$, in which case its probability distribution is centered around $0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.