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Let $X$ be a uniform random variable on $[0,1]$, let $Y$ be uniform in $[3,5]$ independent of $X$. Find the probability density function of $X + Y$.

My solution is: $$ (f_X * f_Y)(x) = \begin{cases} \displaystyle \int_{0}^{x-3} f_X(y) f_Y(x-y)\,dy & \text{if }3<x<4, \\[10pt] \displaystyle\int_{0}^1 f_X(y) f_Y(x-y)\,dy & \text{if }4<x<5, \\[10pt] \displaystyle\int_{x-5}^1 f_X(y) f_Y(x-y)\,dy & \text{if }5< x < 6, \\[10pt] 0 & \text{otherwise}, \end{cases} $$

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You are not finished !!! You have used your knowledge that $f_X$ is zero outside of $[0,1]$. But you also know its value inside $[0,1]$... Same for $f_Y$. Substitute the values and compute the integrals.

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  • $\begingroup$ Ok, yes I know that the integral needs to be computed, but is it true what I wrote? $\endgroup$ – LearningProb Dec 5 '19 at 19:12
  • $\begingroup$ Yes, what you wrote is true, I was just wondering whether you realised that all the information given in the question already allowed you to complete the exercise and give an exact answer ! $\endgroup$ – justt Dec 6 '19 at 17:56

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