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Let $X$ and $Y$ be two variables which are truncated normally distributed above zero (that is $X$ and $Y$ have the lower truncation point zero, their values are bounded above zero). Is $X+Y$ truncated normally distributed as well?

I read all the other questions and answers regarding the sum of truncated normally distributed variables and it seems to me like the sum is not truncated normally distriubted (even if $X$ and $Y$ have the same support). Is this correct?

More importantly, what is the intutions behind? Intuitively, I would actually expect that the sum of $X$ and $Y$ is truncated normally distributed as well. Which problem do I overlook?

Thank you.

Michael Hardy
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Tom
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  • I would say http://math.stackexchange.com/questions/831714/sum-of-two-truncated-gaussian – georg Jul 24 '15 at 11:30

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In an answer to this question, I showed that if $X$ and $Y$ are independent zero-mean normal random variables with the same variance $\sigma^2$, then, with $\hat{X}$ and $\hat{Y}$ being random variables whose joint density is $4f_{X,Y}(x,y)\mathbf 1_{\{x>0, y> 0\}}$, $$F_{\hat{X}+\hat{Y}}(\alpha) = P\{\hat{X}+\hat{Y} \leq \alpha\} = \left[\Phi\left(\frac{\alpha}{\sqrt{2}\sigma}\right) - \Phi\left(\frac{-\alpha}{\sqrt{2}\sigma}\right)\right]^2 = \left[2\Phi\left(\frac{\alpha}{\sqrt{2}\sigma}\right) - 1\right]^2.$$ Can you work out from this whether the sum $\hat{X}+\hat{Y}$ has a truncated normal distribution or not?

As to where your intuition is going wrong, consider whether what your intuition is telling you is the same as insisting that for arbitrary real numbers $x$ and $y$, $ |x| + |y|$ always equals $|x+y|$. The claim is true if $x$ and $y$ are positive numbers ....

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Dilip Sarwate
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