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A very simple question about the variance!

I'm interested in the variance of an expression with a random variable inside it.

For an expression of this form:

$a = \sum_{i=1}^{m}(x_i \cdot y_i - z_i)$

where $y_i$ is a random variable with an expectation $\mathbb{E}(y_i)$, would the expectation of the whole function simply be:

$\mathbb{E}(a)=\sum_{i=1}^{m}(x_i \cdot \mathbb{E}(y_i) - z_i)$

Since variance is $Var(a)=\mathbb{E}(a^2)-(\mathbb{E}(a))^2$, I assume I can simply plug this expression (if correct) into this variance formula. Is this correct?

Thanks!

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If $x_i$ and $z_i$ are deterministic constants for each $i = 1, \ldots, m$, and $m$ is also a known constant--that is to say, $y_i$ for $i = 1, \ldots, n$ are the only random variables involved, and they are independent--then the variance of $a$ is better calculated as $$\operatorname{Var}[a] \overset{\text {ind}}{=} \sum_{i=1}^m x_i^2 \operatorname{Var}[y_i].$$ This is because $z_i$, being deterministic, has zero variance. If the $y_i$ are not independent, then the variance of $a$ will require knowing the covariances between each $y_i$, $y_j$ for $1 \le i \ne j \le m$ in addition to the variances of each $y_i$.

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  • $\begingroup$ Aha, I wasn't aware of the rule about constants being squared in the variance! Thanks. I'm still learning ;) $\endgroup$
    – Sprog
    Dec 31, 2015 at 0:33

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