If a random variable $X$ has mean $\mu_{X}$ and variance $\sigma_{X}^{2}$, and follows a normal distribution, it may be written as

$X\sim \mathcal{N}(\mu_{X},\sigma_{X}^{2})$.

Suppose $Y$ is also normally distributed variables. Should I now write

$X+Y\sim \mathcal{N}(\mu_{X}+\mu_{Y},\sigma_{X}^{2}+\sigma_{Y}^{2})$


$(X+Y)\sim \mathcal{N}(\mu_{X}+\mu_{Y},\sigma_{X}^{2}+\sigma_{Y}^{2})$

or something else? Is there some standard notation for this?

  • $\begingroup$ The mean of the new variable would be the average of $\mu_X$ and $\mu_Y$, not their sum. $\endgroup$ Jul 20, 2015 at 18:30
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    $\begingroup$ Maybe I misread, but the only difference seems to be the parenthesis. This (clearly) doesn't matter. $\endgroup$
    – kasterma
    Jul 20, 2015 at 18:31
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    $\begingroup$ If $X$ and $Y$ are jointly normal, then $X+Y$ is also normal with mean $\mu_X+\mu_Y$ as you state, but the variance of $X+Y$ can be any number ranging from $(\sigma_X-\sigma_Y)^2$ to $(\sigma_X+\sigma_Y)^2$ with the value $\sigma_X^2+\sigma_Y^2$ that you want to use being achieved exactly when $X$ and $Y$ are uncorrelated random variables (including, as a special case, independent random variables). $\endgroup$ Jul 20, 2015 at 20:28
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    $\begingroup$ The $\sim$ notation takes the same precedence as equality. So $x + y \sim \mathcal{N}$ is no more ambiguous than $x + y = z$ and not thinking of $y=z$ as some "operation". They are distinct sides of a certain type of equation ($\sim$ means "has distribution function equal to") $\endgroup$
    – AdamO
    Jul 20, 2015 at 20:35
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    $\begingroup$ Until you modify your question sufficiently to make the claim true you shouldn't write that in either form but the round brackets/parentheses make no difference notationally (i.e. once fixed by including the additional conditions needed to make the assertion true it's fine to either include or omit the parentheses) $\endgroup$
    – Glen_b
    Jul 21, 2015 at 0:39

1 Answer 1


You can certainly write $X + Y \sim N(\mu_{X} + \mu_{Y}, \sigma^{2}_{X} + \sigma^{2}_{Y})$ because if $X$ and $Y$ are normally distributed with the parameters you suggest and are independent, then this statement is true.

However, generally you would write something like $Z = X + Y$, therefore $Z \sim N(\mu_{X} + \mu_{Y}, \sigma^{2}_{X} + \sigma^{2}_{Y})$ so that you can start making statements about, and manipulating, $Z$.


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