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Shocks occur to a system according to a Poisson process of rate $\lambda$. Suppose the system survives each shock with probability $a$, independently of other shocks, so that its probability of surviving $k$ shocks is $ a^{k} $. What is the probability that the system is surviving at time $t$?

My reasoning so far is:
$$\Pr[k \text{ shocks up to time } t] = \Pr[X(t) = k] \frac{(\lambda t)^{k} e^{-\lambda t}}{k!} $$ $$ \Pr[\text{Survive k shocks}] = a^{k} $$ $$\Pr[\text{surviving at time }t] = \Pr[\text{Survive k shocks}] \Pr[k \text{ shocks up to time }t] = \frac{(\lambda t)^{k} e^{-\lambda t}}{k!} a^{k} $$

Please correct me if this reasoning is not correct. Thank you.

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Your reasoning does not appear to be fully correct.

Given that the system underwent $n$ shocks, it survives with probability $p_n = a^n$.

The probability that the system survives at time $t$ is a weighted sum, with weights being the probability that the system was subjected to $n$ shocks: $$ \sum_{n=0}^\infty a^n \Pr\left(N(t)=n\right) = \sum_{n=0}^\infty a^n \frac{(\lambda t)^n}{n!} \exp(-\lambda t) = \exp(a \lambda t) \exp(-\lambda t) = \exp\left((a-1) \lambda t \right) $$

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  • $\begingroup$ This weighted sum perspective makes sense, this made me realize could we also state this in terms of the law of total probability? Pr[surviving at time t] = (sum over n){ Pr[survive n shocks|survive n shocks up to time t] Pr[survive n shocks up to time t]}. $\endgroup$ – rhl Mar 19 '13 at 19:01
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Things look good. Then you need to sum over all $k$. The $e^{-\lambda t}$ is a common factor. For the rest, note that you are finding $\sum \frac{(a\lambda t)^k}{k!}$. Recall that $\sum \frac{x^k}{k!}=e^x$.

Note that the lifetime $T$ of the device (not explicitly asked about) has a familiar distribution.

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