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I had some help on a previous post where I learned that

The distribution of $X(t)-X(s)$, for $s<t$, is Poisson with rate $\lambda(t-s)$. That is, $$ \mathbb P(X(t)-X(s)=k) = > \frac{e^{-\lambda(t-s)}(\lambda(t-s))^k}{k!}.$$

Now the question is as follows:

"Errors in a computer surfaces according to a Poisson process with rate 0.4 per day. Calculate the probability of errors surfacing in three consecutive 5-day periods, with at least 1 error surfacing in each period."

Here, $\lambda=0.4/$day, and so I figured for a 5-day period the rate would be $\lambda=0.4/5$ = $0.08$.

Now when calculating this probability, would it be correct to assume an error occurring in one period is independent of an error occurring next period? and regarding "at least 1 error surfacing in each period", how should that condition be be taken in consideration?

DOGOFWALLSTREET
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1 Answers1

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Here, $\lambda=0.4/$day, and so I figured for a 5-day period the rate would be $\lambda=0.4/5$ = $0.08$.

If you want the rate for the 5-day period it is $0.4\times 5$, not $0.4/5$. You can proceed that way and change the time paramater to 5-day units but it's not necessary: I'll use time units of $1$ day and let $X(t)$ be the number of errors in $t$ days. The arrival rate then remains at $\lambda = 0.4$.

would it be correct to assume an error occurring in one period is independent of an error occurring next period?

Yes.

The probability, $P$, of at least one error in each of $3$ consecutive 5-day periods is:

\begin{eqnarray*} P &=& P(X(5) \geq 1 \; \cap \; X(10) - X(5) \geq 1 \; \cap \; X(15) - X(10) \geq 1) \\ &=& P(X(5) \geq 1)\; P(X(10) - X(5) \geq 1)\; P(X(15) - X(10) \geq 1) \qquad\text{by independent increments} \\ &=& [P(X(5) \geq 1)]^3 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{by stationary increments} \\ &=& [1 - P(X(5) = 0)]^3 \\ &=& \cdots \qquad\qquad\text{hopefully you can finish using the formula you provided and with $\lambda=0.4$.} \end{eqnarray*}

Mick A
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