Question

For random variables X and Y, show that

$\sqrt{\mathrm{Var}(X+Y)}\le \sqrt{\mathrm{Var}\left(X\right)}+\sqrt{\mathrm{Var}\left(Y\right)}$

That is, show that the standard deviation of a sum is always less than or equal to the sum of the standard deviations.

Short answer

The standard deviation is the square root of variance that is$\sqrt{\mathrm{Var}(X+Y)}\le \sqrt{\mathrm{Var}\left(X\right)}+\sqrt{\mathrm{Var}\left(Y\right)}$

Step-by-step solution

Given information

X and Y are random variable

That has the equation$\sqrt{\mathrm{Var}(X+Y)}\le \sqrt{\mathrm{Var}\left(X\right)}+\sqrt{\mathrm{Var}\left(Y\right)}$

Solution 

We need to define the following equations first,

$\mathrm{Var}\left(X\right)={\sigma }_x^2$

$\mathrm{Var}\left(Y\right)={\sigma }_y^2$

The relation of variance and covariance show that,

$\mathrm{Var}(X+Y)={\sigma }_x^2+{\sigma }_y^2+2\mathrm{Cov}(X,Y)$

$\mathrm{Var}(X+Y)={\sigma }_x^2+{\sigma }_y^2+2{\sigma }_x{\sigma }_y$

The property of correlation Shows that,

$\mathrm{Corr}(X,Y)=\frac{\mathrm{Cov}(X,Y)}{{\sigma }_x{\sigma }_y}$

The preceding inequality will becomes,

$\mathrm{Corr}(X,Y)=\frac{\mathrm{Cov}(X,Y)}{{\sigma }_x{\sigma }_y}\le 1$

So,

$\frac{\mathrm{Cov}(X,Y)}{{\sigma }_x{\sigma }_y}\le 1$

$\mathrm{Cov}(X,Y)\le {\sigma }_x{\sigma }_y$

Solution

Now use the above results to obtain the result,

$\mathrm{Var}(X+Y)={\sigma }_x^2+{\sigma }_y^2+2\mathrm{Cov}(X,Y)$

$\mathrm{Var}(X+Y)\le {\sigma }_x^2+{\sigma }_y^2+2{\sigma }_x{\sigma }_y$

$\mathrm{Var}(X+Y)\le {\left({\sigma }_x+{\sigma }_y\right)}^2$

$\mathrm{Var}(X+Y)\le (\sqrt{\mathrm{Var}\left(X\right)}+\sqrt{\mathrm{Var}\left(Y\right)}{)}^2$

Final answer

The standard deviation is the square root of variance that is$\sqrt{\mathrm{Var}(X+Y)}\le \sqrt{\mathrm{Var}\left(X\right)}+\sqrt{\mathrm{Var}\left(Y\right)}$