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Background: According to Wikipedia, the PDF of the sum of two random variables $X$ and $Y$ is given by the convolution: $$f_{X + Y}(x) = \int_{-\infty}^{\infty} f_X(\eta) f_Y(x - \eta) \; d\eta$$ Therefore for the PDF of $X + X$ we have: $$f_{X + X}(x) = \int_{-\infty}^{\infty} f_X(\eta) f_X(x - \eta) \; d\eta$$ Additionally, the stated PDF of the product of two random variables $X$ and $Y$ is given by the integral: $$f_{XY}(x) = \int_{\infty}^{\infty} f_X(\eta) f_Y(x/\eta) \frac{1}{|\eta|} \; d\eta$$ It follows from this formula that the PDF of a constant and a random variable is given by: $$f_{\alpha X}(x) = \int_{-\infty}^{\infty} \delta(\eta - \alpha) f_X(x/\eta) \frac{1}{|\eta|} \; d\eta$$ $$f_{\alpha X}(x) = \frac{f_X(x/\alpha)}{|\alpha|}$$ Example: Suppose we let $X \sim N(0, 1)$ such that the PDF of $X$ is: $$f_X(x) = \frac{1}{\sqrt{2 \pi}}e^{-x^2/2}$$ This gives the PDF of $X + X$ as: $$f_{X + X}(x) = \int_{-\infty}^{\infty} e^{-\left((x - \eta)^2 - \eta^2\right)/2} \; dx$$ $$f_{X + X}(x) = \frac{1}{2\sqrt{\pi}}e^{-x^2/4}$$ Whereas the PDF of $2X$ is: $$f_{2X}(x) = \frac{1}{2}f_X(x/2)= \frac{1}{2\sqrt{2\pi}}e^{-x^2/8}$$ Therefore we see that $X + X$ is distributed as: $$X + X \sim N(0, \sqrt{2})$$ While $2X$ is distributed as: $$2X \sim N(0, 2)$$ The Question: Why is this the case? Clearly the mathematics implies that variables $X + X$ and $2X$ have distinct meanings, however I am struggling to find an intuitive explanation for why they would be different. It seems that for a physical property distributed as $X$ that $X + X$ and $2X$ would be identically distributed. It could also be the case that this is true and that I am somehow misapplying the formulas given above. In either case, I would appreciate any insight.

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$2X$ and $X + X$ are the same random variable. Your confusion arises since you forgot a key condition of the convolution: It gives the distribution or pdf of two independent random variables. Obviously, $X$ is not independent from itself, so you cannot apply the convolution formula in your setting. Additionally, in your example, $2X$ has variance $\sigma^2 = 4$ and it is more common to write this as $N(0,4)$ than $N(0,2)$.

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  • $\begingroup$ I think OP is using the (less common) convention of writing $\mathcal{N}(\text{mean}, \text{standard deviation})$ as opposed to $\mathcal{N}(\text{mean}, \text{variance})$ - that would fit with both the results they derived. $\endgroup$ Commented Oct 14, 2023 at 7:15
  • $\begingroup$ @RubenvanBergen Yes, I clarified that in the post. Thanks. $\endgroup$ Commented Oct 14, 2023 at 7:44
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(Edit: it looks like the answer by PBulls that I referenced below has now been deleted.)

I have to disagree with the answer given by PBulls. I would always interpret $X+X$ s.t. when we have a draw of $X$, the corresponding draw of the sum $X+X$ would be given by that drawn value plus itself, which is the same as twice that drawn value. In other words, $X+X=2X$.

See also this answer from Math Exchange which agrees with me.

In other words, the premise of your question is false: $X+X$ does in fact have the same distribution as $2X$. The convolution of PDF's only applies when adding two independent random variables, $X$ and $Y$. If $X=Y$, then they are not independent - far from it.

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  • $\begingroup$ Yeah, I really didn't approach this with any form of mathematical rigor & seem to have started from the same error as the question - that X and X are independent. I'll stick to more applied questions for a while ;-) $\endgroup$
    – PBulls
    Commented Oct 14, 2023 at 7:49

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