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Let $X, Y$ be iid standard uniform variables, and let $T = X + Y$. The goal is to find the joint PDF of $X$ and $T$. The work I've done so far is to find the PDF of $T$ by evaluating the convolution, but I'm not even sure if that's useful because $X$ and $T$ aren't independent, so the joint PDF shouldn't be the product of the marginal PDFs. I thought about working from the CDF and writing it as $P(X \leq x, T \leq t) = P(X \leq x, X + Y \leq t)$. I didn't know where to go from there due to the lack of independence. Can anyone provide suggestions?

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You are almost there. Note that $$ P(X\le x, T \le t) = P(X \le x, X+Y \le t) = P(X \le x, X \le t - Y), $$ It follows from $$ P(X \le x, X \le t - Y| Y = y) = \int_0^{\min\{x, t- y\}} du = \min\{x, t-y\} $$ that \begin{align*} &P(X \le x, X \le t - Y) \\ &= \int_0^1 P(X \le x, X \le t - Y| Y = y) dy \\ &= \int_0^1 \min\{x, t-y\} dy \\ &= \int_0^{\max\{t-x,0\}} x dy + \int_{\max\{t-x,0\}}^1 (t-y)dy \\ &= x\max\{t-x,0\} + (1-\max\{t-x,0\})t -\frac{1 - (\max\{t-x,0\})^2}{2}. \end{align*}

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  • $\begingroup$ Thank you so much! I definitely follow your explanation. I was wondering if it also might be possible to compute the PDF of X,T via a linear transformation. Since T = X + Y, would it be correct to say that $\begin{bmatrix}1&0\\1&1\end{bmatrix} (X, Y) = (X, T)$? And the matrix has determinant $1$. But then we have that $f_{X,T} = f_{X, Y}(x, t-x}$, which is still $1$ because $f_{X, Y}$ is 1 on the support of $X, Y$. I am sure that something I did is not right?! $\endgroup$ – 0k33 Dec 12 '18 at 21:54

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