# What is the standard deviation of is sigma(X)=2sigma(Y), what is sigma(X+Y)?

Jul 28, 2016

$\sqrt{5} \sigma \left(Y\right)$

#### Explanation:

First note that
$\sigma \left(Z\right) = \sqrt{V a r \left(Z\right)}$
$V a r \left({Z}_{1} + {Z}_{2}\right) = V a r \left({Z}_{1}\right) + V a r \left({Z}_{2}\right)$
$V a r \left[a \cdot Z\right] = {a}^{2} \cdot V a r \left[Z\right]$
because

$V a r \left[{\left(a Z - a \mu\right)}^{2}\right] = V a r \left[{a}^{2} {\left(Z - \mu\right)}^{2}\right] = {a}^{2} V a r \left[{\left(Z - \mu\right)}^{2}\right]$

using expected variance we obtain

$\sqrt{V a r \left(X\right)} = 2 \cdot \sqrt{V a r \left(Y\right)}$

$\sqrt{V a r \left(X\right)} = \sqrt{4 V a r \left(Y\right)}$

$\sqrt{V a r \left(X\right)} = \sqrt{V a r \left(2 Y\right)}$

thus

$\sigma \left(X\right) = \sigma \left(2 Y\right)$

this means every time we see $\sigma \left(X\right)$ we can replace it with
$\sigma \left(2 Y\right)$

$\sigma \left(X + Y\right) = \sqrt{V a r \left(2 Y\right) + V a r \left(Y\right)}$

$\sqrt{V a r \left(2 Y\right) + V a r \left(Y\right)} = \sqrt{4 V a r \left(Y\right) + V a r \left(Y\right)} = \sqrt{5} \sigma \left(Y\right)$