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Let be $X,Y$ two independent uniformly distributed random variables on $[0,1]^2$. Calculate the expected value $\mathbb{E}(X-Y)$.


Actually, it seems pretty easy as we can simply use the properties of the expected value and get $$ \mathbb{E}(X-Y)=\mathbb{E}(X)-\mathbb{E}(Y)=0 $$ because both random variables obey the same distribution. If I define the new random variable $Z:=X-Y$ and compute its probability density function (pdf) by the convolution formula it should yield the same result. We know that the pdfs $f_X(x)=f_Y(y)=1$ for all $0\leq x,y\leq 1$, so

$$ f_Z(z)=\int\limits_{-\infty}^{\infty}f_X(x)f_y(z-x)~dx=\int\limits_{0}^z1\cdot 1 ~dx=z, $$ The range of $Z$ is $[-1,1]$ so we get $$ \mathbb{E}(Z)=\int\limits_{-1}^1z\cdot z ~dz=\frac{z^3}{3}\Big|_{-1}^1=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\neq 0. $$

This contradicts the result from the beginning. Where is my mistake?

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    $\begingroup$ Are you saying that the probability density of $Z$ is $f(z)=z$? But that's not $≥0$ for all $z$ and it doesn't integrate to $1$ over the range of $Z$. $\endgroup$
    – lulu
    Aug 15, 2023 at 13:00
  • $\begingroup$ @lulu Ah ok I see a first mistake: the pdf of $Z$ must be $0$ if $-1\leq z<0$. But I don't see the mistake that prevents integrating to $1$ over the range of $Z$. I must have evaluated the integral wrong but I don't see where? $\endgroup$
    – Philipp
    Aug 15, 2023 at 13:23
  • $\begingroup$ here is a near duplicate. $\endgroup$
    – lulu
    Aug 15, 2023 at 13:26

1 Answer 1

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You are convolving $X$ and $-Y$ but you are using the formula for $X+Y$ instead of $X-Y$. In particular, you should use the pdf of $-Y$ which is supported on $[-1,0]$ instead of that of $Y$. So infact, the range of integration will be from $z$ to $1$ and you'll get $1-z$ when $z$ is positive and it will be from $0$ to $1+z$ when $z$ is negative. So $f(z)=\begin{cases} (1+z)\,,-1\leq z\leq 0\\ (1-z)\,, 0\leq z\leq 1\end{cases}$

Now compute the expectation and you'll get $0$

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