Let $(X,Y)$ be uniform $(0, 1) \times (0, 1)$ random vector and $Z=\min \{X,Y\}$. Find $M(1)$

Question

Let $(X,Y)$ be uniform $(0, 1) \times (0, 1)$ random vector and $Z=\min \{X,Y\}$. Find $M(1)$, where $M(t)$ is the moment generating function.

I do calculate and my answer is $2(e-1)$. Can someone check please?

Answer 1

let $z \in (0,1)$, $$Pr(Z \le z) = 1 - Pr(Z > z)=1-Pr(X>z)Pr(Y>z)=1-\left( 1-z\right)^2$$

$$f_Z(z)=\begin{cases} 2(1-z) & ,z \in (0,1) \\ 0&, z \notin (0,1) \end{cases}$$

$$M_Z(t)=E(\exp(tZ))=\int_0^12(1-z)\exp(tz) \, dz$$

\begin{align}M_Z(1)&=2\left(\int_0^1 \exp(z)\, dz-\int_0^1 z\exp(z) \, dz\right) \\ &= 2\left(\int_0^1 \exp(z)\, dz-z\exp(z)|_0^1+\int_0^1 \exp(z) \, dz\right) \\ &=2\left(2\int_0^1 \exp(z)\, dz-z\exp(z)|_0^1\right)\\ &=2\left(2(\exp(1)-1)-\exp(1)\right)\\ &=2(\exp(1)-2)\\ &=2(e-2)\end{align}