I want to prove that $$\operatorname{Var}\left(\sum\limits_{i=1}^m{X_i}\right) \leq m\sum\limits_{i=1}^m{\operatorname{Var}(X_i)} \,. \>$$

A too complicated proof is to write $$ a_{ij}=\sqrt {Cov(X_i,X_j)} \,, $$

$$\operatorname{Var}\left(\sum\limits_{i=1}^m{X_i}\right) = \sum a_{ij}^2 \leq \sum a_{ii}a_{jj} \leq \sum\sum a_{ii}^2$$

By Cauchy-Scwarz and then the permutation inequality.

I'm sure it can be shorter, but how?

  • 3
    $\begingroup$ Are there any assumptions on $X_i$? If $a_{ij} = \sqrt{Cov(X_i, X_j)}$ then you are implicitly assuming that the covariance is non-negative. Is this given? $\endgroup$ Commented Jul 25, 2017 at 7:54
  • $\begingroup$ @Greenparker: given the $\operatorname{Var}(X_i)$, the variance of the sum will be maximised when all the correlations are $+1$. You might still need to show something like $\left(\sum\limits_{i=1}^m{k_i}\right)^2 \le m \sum\limits_{i=1}^m{k_i^2 }$ $\endgroup$
    – Henry
    Commented Jul 25, 2017 at 8:45
  • $\begingroup$ Expand $\text{Var}(\sum X_i)$ in terms of variances and covariances. Write $\text{Cov}(X_i,X_j)$ in terms of standard deviations of the $i$th and $j$th $X$ and their correlation. Use bounds on correlation (which itself can be shown in a variety of ways if you can't just take it as given). $\endgroup$
    – Glen_b
    Commented Jul 25, 2017 at 11:39

2 Answers 2


The inequality is true in general under the assumption $\textrm{Var}(X_i)<\infty$ for all $i$ (otherwise both sides are infinite). Applying Cauchy-Schwarz to the random quantities $X_i-\mathbb E X_i$ yields that $$ \mathbb P\left[\left(\sum_{i=1}^m X_i-\mathbb E X_i\right)^2\leq m\sum_{i=1}^m (X_i-\mathbb E X_i)^2\right]=1. $$ Consequently, $$ \mathbb E\left(\sum_{i=1}^m X_i-\mathbb EX_i\right)^2\leq m\ \mathbb E\sum_{i=1}^m (X_i-\mathbb EX_i)^2. $$ Using the definition of variance, we see this is exactly the desired inequality.


A simpler method is to use the inequality $a_ia_j \leq \frac 12 (a_i^2+a_j^2)$ after Cauchy-Schwarz.


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