$X\thicksim\text{Uniform}[0,2]$ and $Y\thicksim\text{Exp}(\lambda)$ are independent, find $P(X<Y)$ and $F_{\min(X,Y)}(z)$

Question

The Problem Let $X\thicksim\text{Uniform}[0,2]$ and $Y\thicksim\text{Exp}(\lambda)$. Assume that $X$ and $Y$ are independent.
(a) Find the probability $P(X<Y)$.
(b) FInd the CDF of $Z=\min(X,Y)$. Check whether $Z$ is absolutely continuous, and find its PDF if it has one.

Thoughts:
(a) Recall that $$f_X(x)=\begin{cases}\dfrac{1}{2}&\text{if }0\leq x\leq2\\0&\text{otherwise}\end{cases}\quad\text{and}\quad f_Y(y)=\begin{cases}\lambda e^{-\lambda y}&\text{if }y\geq0\\0&\text{otherwise.}\end{cases}$$ Now we find the joint PDF of $(X,Y)$, which by independence is given by $$f_{XY}(x,y)=\begin{cases}\dfrac{\lambda e^{-\lambda y}}{2}&\text{if }0\leq x\leq2,y\geq0\\0&\text{otherwise.}\end{cases}$$ Then the probability in question can be found by integrating the joint PDF over the region $D=\{(x,y)\,:\,x<y\}.$ We have \begin{align*} P(X<Y)&=\iint\limits_Df_{XY}(x,y)\,dy\,dx=\int_0^2\int_{x}^{\infty} f_{XY}(x,y)\,dy\,dx\\ &=\int_0^2\int_{x}^{\infty}\frac{\lambda e^{-\lambda y}}{2}\,dy\,dx=\int_0^2\frac{1}{2}e^{-\lambda x}\,dx\\ &=\frac{1}{2\lambda}\left[1-e^{-2\lambda}\right]. \end{align*}

(b) Using independence we have that \begin{align*} P(Z\leq z)&=P(\min(X,Y)\leq z)=1-P(X>z,Y>z)\\ &=1-P(X>z)\cdot P(Y>z)\\ &=1-[1-F_X(z)]\cdot[1-F_Y(z)]. \end{align*} Therefore we have the following case-defined CDF $$F_Z(z)=\begin{cases}0&\text{if }z<0\\ 1-e^{-\lambda z}\left[1-\dfrac{z}{2}\right]&\text{if }0\leq z\leq2\\ 1&\text{if }z>2.\end{cases}$$ The CDF is continuous everywhere and differentiable almost everywhere, so we may differentiate it to obtain the PDF $$f_Z(z)=\begin{cases}\lambda e^{-\lambda z}+\dfrac{e^{-\lambda z}}{2}-\dfrac{\lambda ze^{-\lambda z}}{2}&\text{if }0\leq z\leq2\\0&\text{otherwise.}\end{cases}$$

Therefore, $Z$ is absolutely continuous.

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Do you agree with my work above? Any comments are most welcome and highly appreciated.
Thank you very much for your time.

Answer 1

Excellent! Especially the way you solved b) pleases me.

Alternative for a):$$\begin{aligned}P\left(X<Y\right) & =\int P\left(X<Y\mid X=x\right)f_{X}\left(x\right)dx\\ & =\frac{1}{2}\int_{0}^{2}P\left(x<Y\mid X=x\right)dx\\ & =\frac{1}{2}\int_{0}^{2}P\left(x<Y\right)dx\\ & =\frac{1}{2}\int_{0}^{2}e^{-\lambda x}dx\\ & =\frac{1}{2}\left[-\frac{e^{-\lambda x}}{\lambda}\right]_{0}^{2}\\ & =\frac{1-e^{-2\lambda}}{2\lambda} \end{aligned} $$ where the third equality rests on independence.