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