Min. of a sequence of r.v's with Poisson index

Let there be a sequence of r.v.'s $\{X_{n}\}_{n\geq1}$ each with distribution $U(0,1)$ and another sequence $N_{k} \sim \operatorname{Poisson}(k)$, defining a new sequence as \begin{align} Y_{k} = \left\{ \begin{array}{ll} \; &0 \;&\text{if} \;N_{k}=0 \\&k \min\{X_{1},...,X_{N_{k}}\} \;&\text{if} \;N_{k}\geq1 \end{array}\right. \end{align}

Find if $Y_{k} \xrightarrow[]{D} Y$ and Y expression.

We know that the $U(0,1)$ and the minimum probability when conditioned are

$F_{x_{n}}=\int_{0}^{a}dx=a$

$F_{\min\{X_{1},..,X_{N_{k}}\}|N_{k}=n}=1-(1-F_{X_{n}})^{n}$

And to find the distribution of $Y_{n}$, we need to sum over all possible n's and their probabilities to "decondition" it, \begin{align} &\operatorname{P}(\min\{X_{1},..,X_{N_{k}}\}\leq \frac{a}{k})\\ =&\sum_{n=0}^{\infty}\operatorname{P}(\min\{X_{1},..,X_{N_{k}}\}\leq \frac{a}{k}|N_{k}=n)\operatorname{P}(N_{k}=n)\\ =&\sum_{n=0}^{\infty}(1-(1-\frac{a}{k})^{n})e^{-k}\frac{k^{n}}{n!} \end{align} And this is where I bump heads with the problem, i know that $\sum_{n=0}^{\infty}\frac{k^{n}}{n!}=e^{k}$ and i must use this property to develop my results, however I've tryed many ways and must missed something along the way.

My book says that $F_{X_{n}} = 1+e^{-k}-e^{a}$, which clearly converges to a Exp(1) when $k \xrightarrow[]{} \infty$, I'm just not sure how to get to this expression. I'd very much appreciate help, thanks in advance.

Solution 1:

Note that when $N_k = 0$ we have $Y_k = 0$ which is smaller than any $a>0$ with probability one.

The target CDF $F_{X_n}$ should really be denoted as $F_{Y_k}$ or $F_{Y_n}$, but I shall adhere to the source. \begin{align*} F_{X_n}(a) &\equiv \Pr\bigl\{ Y_k < a\bigr\} \\ &= 1 \cdot \Pr\bigl\{ N_k = 0\bigr\} + \sum_{n = 1}^{\infty} \Pr\bigl\{ k\cdot \min\{X_{1},..,X_{N_{k}} \} < a\bigr\} \cdot \Pr\bigl\{ N_k =n \bigr\} \\ &= e^{-k} + \sum_{n = 1}^{\infty} \Pr\left\{ \min\{X_1,..,X_n \} < \frac{a}k \right\} e^{-k} \frac{ k^n }{ n!} \\ \end{align*}

Where this summation is the same as your "de-conditioning" equations, except that the index starts at one: \begin{align*} \text{summation} &= e^{-k}\sum_{n=1}^{\infty}\left( 1- \left( 1-\frac{a}k \right)^n \right)\frac{k^{n}}{n!} \quad\qquad\text{, now split the sum}\\ &= e^{-k} \left[ \sum_{n=1}^{\infty}\frac{k^n}{n!} - \sum_{n=1}^{\infty}\left( 1- \frac{a}k \right)^n \frac{k^{n}}{n!} \right]\\ &= e^{-k} \left[ e^k - 1 - \sum_{n=1}^{\infty}\frac{ \bigl( k ( 1- a/k ) \bigr)^n }{n!} \right] \\ &= e^{-k} \bigl[ e^k - 1 - \left( e^{k - a} - 1\right) \bigr] \\ &= 1 - e^{-a} \end{align*} Put things back together, we arrive at the result provided by the book. $$F_{X_{n}}(a) = e^{-k} + \text{summation} = 1 +e^{-k} - e^{-a}$$