Conditional expectation $E(Z_1+...+Z_N | \min(Z_1,...,Z_N) )$

Question

Let $Z_1,Z_2,...$ be i.i.d. random variables with density: $$f(x)=\frac{\lambda^{\theta} \theta }{(\lambda+x)^{\theta+1}} \mbox{ for } x>0,$$ where $\theta>1, \lambda>0$. Let $N$ be independent of $Z_1,Z_2,..$ with geometric distibution $$P(N=n)=(1-q)q^{n-1} \mbox { for } n=1,2,...$$ where $q \in (0,1)$. $$E(Z_1+...+Z_N | \min(Z_1,...,Z_N)=t)=?$$ I thought that it would be possible to use tower property to introduce $N$ as a condition and then use Bayes' theorem, but I'm not sure if there is a proper inclusion of $\sigma$-algebras to use the tower property. Any suggestions?

Answer 1

You can always condition on anything you want, in this example we have: $$ E[X|\min=t] = E[E[X|N, \min=t]|\min=t] = \sum_{n=1}^{\infty}E[X|N=n,\min=t]Pr[N=n|\min=t] $$ and you can work the problem from this formula.

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Three ways to understand this formula:

1) For a general random variable $X$, we have the "tower property" (also called "iterated expectations"): $$ E[X] = E[E[X|N]] = \sum_{n=1}^{\infty}E[X|N=n]Pr[N=n] $$ Now, if we live in a world where $\min=t$ then everything in the above formula should be conditioned on "$\min=t$."

2) In terms of sigma-algebras, the tower property is $E[X|\mathcal{G}_1] = E[E[X|\mathcal{G}_2]|\mathcal{G}_1]$ whenever $\mathcal{G}_1 \subseteq \mathcal{G}_2$. In this example we have $\mathcal{G}_1 = \sigma(\min)$ and $\mathcal{G}_2 = \sigma(\min, N)$.

3) Define $1\{N=n\}$ as an indicator function that is 1 if $N=n$, and 0 else. Then with prob 1 we have: $$ 1 = \sum_{n=1}^{\infty} 1\{N=n\}$$ Thus, with prob 1: $$ X = \sum_{n=1}^{\infty} X \cdot 1\{N=n\}$$ Now take conditional expectations of both sides, given $min=t$: \begin{align} E[X|\min=t] &= \sum_{n=1}^{\infty} E[X \cdot 1\{N=n\} | \min=t] \\ &= \sum_{n=1}^{\infty} E[X|N=n, \min=t]Pr[N=n|\min=t] \end{align}