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Mixed Distribution Function

$$ F(t) = \begin{cases} \hfill 0 \hfill & t < 0 \\ \hfill p+(1-p)(1-e^{-yt}) & t \geq 0 \end{cases} $$

How can i find the average waiting time of an arrival and average waiting time for an arrival given that a wait is required.

For part 1: I did this but was struck as the equation is tending to infinity

$$ E(t) = \int_{t=0}^\infty t[p+(1-p)(1-e^{-yt})] \, dt $$

But this value is tending to infinity when we do integration by part

$$ = \int_{t=0}^\infty tp \, dt + \cdots $$

The first value is infinity?? Kindly give me some suggestions on this first part. I also don't know what to do with the second one as in my knowledge should be same

Any help is appreciated. Thank you

Brian Tung
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Jake
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1 Answers1

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The function $F(t)$ is a CDF (cumulative distribution function), not a PDF (probability distribution function). It gives

$$ F(t) = P(T \leq t) $$

To determine the average from a CDF, you do not multiply by $t$; you just subtract from $1$ and integrate:

$$ E(T) = \int_{t=0}^\infty [1-F(t)] \, dt $$

The difference between the two questions is whether you include that impulse of size $p$ at the origin. If you do not, then the integral is the same, but you should scale upward by the reciprocal of the probability that you have to wait, or $1/(1-p)$. (That is, you wait longer on average, given that you wait at all.)

Brian Tung
  • 34,160