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$$U \sim \operatorname{unif}(0,1);X \sim \operatorname{expo}(1) .$$ $U,X$ indep.

Find the PDF of $U+X$.

Here's my answer, but I find this answer implausible, not sure where I went wrong.


\begin{align} F_Y(y) & =P(U+X \leq y) \\ &= \int_{-\infty}^{\infty} P(U+X < y \mid X=x)f_X(x)\,dx \\ &= \int_{-\infty}^{\infty} P(U < y - X\mid X=x)f_X(x)\,dx \\ &= \int_{-\infty}^{\infty} P(U < y - x\mid X=x)f_X(x)\,dx \\ &= \int_{-\infty}^{\infty} P(U < y - x)f_X(x)\,dx \\ &= \int_{-\infty}^{\infty} F_U(y-x)f_X(x)\,dx \\ &= \int_{-\infty}^{\infty} (y-x)f_X(x)\,dx \\ &= \operatorname E_X[y-X] \\ &= y - \operatorname E_X[X] \\ &= y - 1/\lambda \\ &= y - 1/1 \\ &= y - 1 \end{align}

$$ f_Y(y) = \frac d {dy} F_Y(y) = 1 $$

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  • $\begingroup$ You seem to have two instances of two identical lines adjacent to each other in this string of equalities. $\endgroup$ – Michael Hardy Dec 10 '17 at 4:16
  • $\begingroup$ @MichaelHardy Thanks, fixed. $\endgroup$ – Hatshepsut Dec 10 '17 at 4:20
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Note that $F_U(y-x)=\begin{cases} 0 & , y-x < 0\\y-x & ,y-x \in [0,1] \\ 1 &, y-x > 1\end{cases}$.

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  • $\begingroup$ Isn't $F_U (y-x) = 1$ when $y-x>1$? $\endgroup$ – Hatshepsut Dec 10 '17 at 4:13
  • $\begingroup$ you are right, i miss that. $\endgroup$ – Siong Thye Goh Dec 10 '17 at 4:38

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