Find the expected value of the following problem.. Let $X_i$, for i=1,2,.... be independent uniform(0,1) random variables. Define the integer valued random variable N which has the following PMF $$f_N(n)=\frac{c}{n!}\quad n\in\{1,2,.....\}$$ where $c=\frac{1}{e-1}$. Derive the probability density function of $T=min_{1\le i\le N}(X_i)$ and use it to find $E(T)$. 
At first I tried to find the CDF of T and then pdf of T which I get $N(1-x)^{N-1}$ that I don't think is right. I am stuck here....
I would highly appreciate if anyone could help me. Thanks in advance...
 A: CDF of $T$ is given by
$$
F(x)=P[T\le x]=\sum_{n=1}^\infty P[X_1\le x\lor X_2\le x\lor\cdots\lor X_n\le x ]P[N=n]\\
=\sum_{n=1}^\infty{c\over n!}\left[1-P[X_1> x\land X_2> x\land\cdots\land X_n> x ]\right]=\sum_{n=1}^\infty{c\over n!}\left[1-P[X_j>x]^n\right]\\
=\sum_{n=1}^\infty{c\over n!}\left[1-\left[\int_{x}^1dt\right]^n\right]=\sum_{n=1}^\infty{c\over n!}[1-(1-x)^n]=1-c(e^{1-x}-1)
$$
PDF of $T$ is given by
$$
f(x)=ce^{1-x}
$$
Expectation of $T$ is given by
$$
\int_{0}^1cxe^{1-x}dx={e-2\over e-1}
$$
A: $\begin{eqnarray}
&&F_T(t)\\&=&P(T\le t)\\&=&\sum_{k=1}^\infty P(T\le t|N=k)P(N=k)\\&=&\sum_{k=1}^\infty P(\min\{X_1,X_2,...,X_k\}\le t)P(N=k)\\&=&\sum_{k=1}^\infty P(\min\{X_1,X_2,...,X_k\}\le t)\frac{c}{k!}\\
&=&c\sum_{k=1}^\infty\frac{P(\min\{X_1,X_2,...,X_k\}\le t)}{k!}\tag1
\end{eqnarray}$
On the other hand, 
$\begin{eqnarray}P(\min\{X_1,X_2,...,X_k\}\le t)&=&1-P(\min\{X_1,X_2,...,X_k\}> t)\\&=&1-P(X_i>t\mbox{ for all }i\in\{1,...,k\})\\
&=&1-P(X_1>t)...P(X_k>t)\\
&=&1-P(X>t)^k\\
&=&1-(1-P(X\le t))^k
\tag2\end{eqnarray}$
Substituying $(2)$ in $(1)$:
$\begin{eqnarray}F_T(t)&=&c\sum_{k=1}^\infty\frac{1-(1-P(X\le t))^k}{k!}\\&=&c\sum_{k=1}^\infty\frac{1}{k!}-c\sum_{k=1}^\infty\frac{(1-P(X\le t))^k}{k!}\\&=&1-c\sum_{k=1}^\infty\frac{(1-P(X\le t))^k}{k!}\\&=&1-c(e^{1-P(X\le t)}-1)\\&=&1-c(e^{1-F_X(t)}-1)\end{eqnarray}$ 
Thus $f_T(t)=cf_X(t)e^{1-F_X(t)}$.
Can you finish?
