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Let $T = X_1 + X_2 + ...+ X_N$, where $X_i \sim (iid) Exp(\lambda) $ and $N \sim Geom(p)$, such that $P(N=k) = p(1 - p)^{k-1}, k=1, 2, ....,$ and $N$ and $X_i$ are independent $\forall i$. Find the distribution of $T$.

I am completely lost with this question, any help is really really appreciated! I don't even know what to look up :(

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1 Answer 1

up vote 2 down vote accepted

You can decompose the problem this way:

$$P[T \leq x] = \sum_{k=1}^{+\infty} P[T \leq x | N=k] P[N=k]$$

$$= \sum_{k=1}^{+\infty} P[ X_1 + ... + X_k \leq x] p(1-p)^{k-1}$$

The sum of the k exponential variables has a gamma distribution.

Working with probability distribution functions:

$$t(x) = \frac{d}{dx} P[T \leq x]$$

$$ = \sum_{k=1}^{+\infty} \frac{d}{dx} P[ X_1 + ... + X_k \leq x] p(1-p)^{k-1}$$

$$ = \sum_{k=1}^{+\infty} \frac{\lambda^k}{\Gamma(k)} x^{k-1} e^{-\lambda x} p(1-p)^{k-1}$$

$$ = \sum_{k=1}^{+\infty} \frac{\lambda^k}{(k-1)!} x^{k-1} e^{-\lambda x} p(1-p)^{k-1}$$

$$ = \lambda \sum_{k=1}^{+\infty} \frac{\lambda^{k-1}}{(k-1)!} x^{k-1} e^{-\lambda x} p(1-p)^{k-1}$$

$$ = p\lambda e^{\lambda x (1-p)} e^{-\lambda x} $$

$$ = p\lambda e^{-\lambda p x} $$

This is therefore another exponential random variable, with parameter $\lambda p$.

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I may need things explained a bit more :/ It's just a random exam paper question but I can't find anything in my notes that seems relevant or to explain what to do.. Could you go step-by-step? Thanks for the help! :) –  Swayy Feb 17 '13 at 21:25
    
I added some steps, tell me if you need more –  GHL Feb 17 '13 at 21:41
    
That's great, thanks so much! –  Swayy Feb 17 '13 at 23:14

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