# If $X$ is a random variable, why is the PDF of $X + X$ not the same as the PDF of $2X$?

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.

$$2X$$ 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 $$N(0,2)$$.

• 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. Oct 14, 2023 at 7:15
• @RubenvanBergen Yes, I clarified that in the post. Thanks. Oct 14, 2023 at 7:44

(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, $$X+X=2X$$.

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.