23
votes
Is the variance of the mean of a set of possibly dependent random variables less than the average of their respective variances?
Yes, it is true. Here is a proof. $$
\begin{align}
\newcommand{\Var}{\operatorname{Var}}
&\Var(\overline{X})
\\
&= \frac1{n^2}\Var\left(\sum_{i=1}^n X_i\right)
\\
&=\frac1{n^2}\sum_{i=1}^n\...
- 78.3k
14
votes
Accepted
Is the variance of the mean of a set of possibly dependent random variables less than the average of their respective variances?
In general, one has :
$$
\begin{align}
\operatorname{Var}\left(\sum_{k=0}^n X_k\right)
&= \sum_{i,j=0}^n \operatorname{Cov}(X_i,X_j)
\end{align}
$$
Now, the well-known inequality $ab \le \frac{...
- 10.3k
13
votes
Is the variance of the mean of a set of possibly dependent random variables less than the average of their respective variances?
A way to see this at a glance is that real random variables form an inner product space, with $\langle X,Y \rangle = \mathbb{E}XY$.
The norm induced by this inner product is $\|X\|^2=\mathbb{E}X^2$, ...
- 5,633
4
votes
Is the variance of the mean of a set of possibly dependent random variables less than the average of their respective variances?
$$\text{Var}\left(\sum X_i\right) = \sum\limits_i \text{Var}\left( X_i\right) +\sum\limits_i \sum\limits_{j\not=i} \text{Cov}\left( X_i,X_j\right)$$ is maximised when the covariances take their ...
- 159k
1
vote
Confusion in using applying variance formula
The variance of a sum of independent random variables $X$ and $Y$ is the sum of their variances; i.e.,
$$\operatorname{Var}[X+Y] \overset{\text{ind}}{=} \operatorname{Var}[X] + \operatorname{Var}[Y],$$...
- 141k
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