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I've been fighting with following problem:

Problem: Let $N, X_1, X_2, \ldots $ be independent random variables with given $\Lambda = \lambda$. Variables $X_i, i=1,2,\ldots$ have exponential distribution with expected value $\mathbb{E}=\frac{1}{\lambda}$ where $\Lambda=\lambda$. Conditional distribution of $N$ with given $\Lambda=\lambda$ is a Poisson distribution with $\mathbb{E}=\lambda$. Marginal distribution of variable $\Lambda$ is a Gamma distribution with density function given by: $$f(\lambda) = \frac{8}{3}\lambda^3e^{-2\lambda}\mathbb{1}_{\lambda > 0}$$

Let $ S = \sum_{i=1}^{N} X_i,\ \ \ N>0$, $S=0$ when $N=0$ and let $T=\sum_{i=1}^{N} Y_i, \ \ N>0$, $Y=0$ when $N=0$, where $Y_i=\min\lbrace X_i, 2\rbrace$.

Find $Cov(S, T)$.

My solution:

We know that:

  1. $Cov(S,T) = Cov[\mathbb{E}(S|N), \mathbb{E}(T|N)] + \mathbb{E}[Cov(S,T)|N]$.

  2. $\mathbb{\ E}\Lambda = \int_0^{\infty} \frac{8}{3} \lambda^4e^{-2\lambda}d\lambda = 2 $.

  3. $\mathbb{E}(S|N) = \mathbb{E}(X_1 + X_2 + \ldots + X_N) = N \cdot \mathbb{E}X_i = N \cdot \mathbb{E}(X_i|\Lambda=\lambda)\cdot\mathbb{E}(\Lambda) = N \cdot \frac{1}{\lambda}\cdot 2 = \frac{2N}{\lambda}$

  4. $\mathbb{E}(T|N) = \mathbb{E}(Y_1 + \cdots + Y_N) = N \cdot \mathbb{E}Y_i = N \cdot \mathbb{E}(\min\lbrace X_i, 2\rbrace) = $ and here I am stuck.

My question is: is this approach correct? I would also appreciate any hints what to do next.

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  • $\begingroup$ Use the law of total probability to condition your expectation on $X \le 2$ and $X > 2$ $\endgroup$ – user365239 Oct 24 '17 at 0:47

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