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Let $(U_n)_n$ a sequence of random variables i.i.d $U[0,1]$ and let $P\sim \mathrm{Poi}(\lambda)$ a random variable such that $P$ is independent of $(U_n)_n$. Let $$ \\ X=\left\{\begin{matrix} \operatorname{min}\{U_1,...,U_P\}, P\ne 0\\ 1,P=0 \end{matrix}\right. $$ Find $F_X(t)$.

I saw that the solution is $$ \ F_X(t)=\left\{\begin{matrix} 0,t\leq 0\\ 1-e^{-\lambda t},0\leq t<1 \\ 1, 1\leq t \end{matrix}\right.\\ $$ But my solution is $F_X(t)=e^{-\lambda}(e^{\lambda t}-1)$ and I don't know where is my mistake:

Obviously for $1\leq t$ then $\mathbb{P}(X\leq t)=1$ and if $t<0$ then $\mathbb{P}(X\leq t)=0$. For $0\leq t <1$, $$ \\ \mathbb{P}(X\leq t)=\mathbb{P}(X\leq t,P=0)+\mathbb{P}(X\leq t,P>0) \ $$ But if $P=0$ then $X=1$ then $\mathbb{P}(X\leq t,P=0)=0$, thus $$ \\ \mathbb{P}(X\leq t)=\mathbb{P}(X\leq t, P>0)=\sum_{k=1}^\infty \left(\prod_{j=1}^k\mathbb{P}(U_j\leq t)\right)\mathbb{P}(P=k)=\sum_{k=1}^\infty {(t\lambda)^k\over k!}\cdot e^{-\lambda} \ \\ =e^{-\lambda}\sum_{k=0}^\infty{(t\lambda)^k\over k!}-e^{-\lambda}= e^{-\lambda}\cdot e^{t\lambda} -e^{-\lambda} =e^{-\lambda}(e^{\lambda t}-1).\ $$

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    $\begingroup$ You seme to assume that $P=k$ implies $\min (U_1,U_2,...,U_P)=U_k$ which is false. $\endgroup$ – Kavi Rama Murthy Jan 20 at 11:45
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Since $P(0\le X\le 1)=1$, it suffices to show that $$ \Bbb P(X>t)=e^{-\lambda t} $$ for all $t\in [0,1)$. Note that $$\begin{eqnarray} \Bbb P(X>t)&=&\sum_{k=0}^\infty \Bbb P(X>t, P=k)\\ &=&\sum_{k=0}^\infty \Bbb P(\min\{U_1,\ldots,U_k\}>t, P=k)\\ &=&\sum_{k=0}^\infty \Bbb P(\min\{U_1,\ldots,U_k\}>t)\Bbb P( P=k)\\ &=&\sum_{k=0}^\infty \Bbb P(U_i>t,\forall i\le k)\Bbb P( P=k)\\ &=&\sum_{k=0}^\infty (1-t)^k\Bbb P( P=k)\\ &=&\sum_{k=0}^\infty (1-t)^k\frac{\lambda^ke^{-\lambda}}{k!}\\ &=&\sum_{k=0}^\infty \frac{(\lambda(1-t))^k e^{-\lambda}}{k!}=e^{\lambda(1-t)}e^{-\lambda}=e^{-\lambda t} \end{eqnarray}$$ as desired.

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