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Convolution of a normal distribution says: If, $X \sim \mathcal{N}(\mu, \sigma^2)$, then $X+X\sim\mathcal{N}(\mu+\mu, \sigma^2+\sigma^2)=\mathcal{N}(2\mu,2\sigma^2)$

However, Multiplication of a random variable with constant says that it also holds that:

$E[c\cdot X]=c\cdot E[X]$ which is fine but additionally

$Var[c\cdot X]=c^2\cdot Var[X]$

or in a different notation: $cX\sim \mathcal{N}(c\mu, c^2\sigma^2)$

There is an additional $^2$ for $\sigma$.

So, is $X+X\neq 2X$? If yes, what is the intuition of $X+X$ and what is the intuition of $2X$?

Community
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IceFire
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2 Answers2

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If $X \sim N(\mu_X ,\sigma_X)$ and $Y \sim N(\mu_Y ,\sigma_Y)$ then $$X + Y \sim N(\mu_X + \mu_Y,\; \sigma_X^2 + \sigma_Y^2 + 2\sigma_{X,Y})$$ We have $$X + X \sim N(\mu_X + \mu_X,\; \sigma_X^2 + \sigma_X^2 + 2\sigma_X^2)$$ Indeed $$2X \sim N(2\mu_X ,\; 4\sigma_X^2 )$$ Note

$$2\sigma_{X,Y}=2\operatorname{Cov}(X,Y) $$ $$2\sigma_{X,X}=2\operatorname{Cov}(X,X)=2\operatorname{Var}(X)=2\sigma_X^2 $$

Behrouz Maleki
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The variable $X$ is not independent of variable $X$. Thus the correlation coefficient is unequal to zero.

The variance of the sum of two dependent, normally distributed random variables (X,Y) is

$\sigma_{X+Y}^2=\sigma_X^2+\sigma_Y^2+2\rho \sigma_X\cdot \sigma_Y$

Setting Y equal to X.

$\sigma_{X+X}^2=\sigma_X^2+\sigma_X^2+2\rho_X \cdot \sigma_X\cdot \sigma_X$

Note that $\rho_X=Corr(X,X)=1$

$\sigma_{X+X}^2=\sigma_X^2+\sigma_X^2+2\cdot 1 \cdot \sigma_X\cdot \sigma_X=4\sigma_X^2$

callculus42
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    Thank you, as well!! Behrouz was faster, which is why I gave him the mark but your solution is equally helpful. – IceFire Jul 20 '16 at 12:55
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    @IceFire You´re welcome. That´s fair, seriously. The main thing is that both answers helps. I appreciate that you have written a comment. – callculus42 Jul 20 '16 at 13:01