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A device randomly beeps, with an average of 2 beeps per second.

If we assume that the time $T$ between any two beeps is an exponential random variable, then what is the probability that no beep is emitted in the next 5 seconds? Now, if we let $Y_i$ be exponential random variables modeling the time between the $i$th and the $i$ + 1st beeps, what is the probability that there are exactly 3 beeps in the next 4 seconds? I know we can parametrize this with a parameter $\lambda$ which is 2 beeps per second, but I am unsure of how to proceed in each of these situations.

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    $\begingroup$ The first question is simply $P(T > 5)$. (Why?) If you are able to use Poisson processes, the second question is the probability that a Poisson random variable (with mean $2 \cdot 4$) equals $3$. $\endgroup$
    – angryavian
    Nov 7, 2019 at 21:04
  • $\begingroup$ We could use a Poisson process, but I am specifically interested in modeling the second case with exponential random variables. $\endgroup$
    – user581882
    Nov 7, 2019 at 21:06
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    $\begingroup$ @theta: Just thinking aloud here for the second part. You need the probability $Y_0+Y_1+Y_2\leq 4$ and $Y_0+Y_1+Y_2+Y_3>4$ where $Y_i$ are exponential RV's with $\lambda=2$. The issue is that these two events don't seem independent, so you can't just multiply probabilities of individual events (which can be calculated via gamma distribution). The second part seems particularly well-suited to use of Poisson distribution. $\endgroup$ Nov 7, 2019 at 21:49

1 Answer 1

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Preliminary

Gamma distribution to the rescue! If a R.V. $T\sim Gamma(a,\mu)$, its pdf is...

$$ P[T = t] = \begin{cases} \frac{t^{a-1}e^{-\frac{t}{\mu}}}{\mu^a\Gamma(a)}, & t \gt 0 \\ 0, & t \le 0 \end{cases} $$

And its cdf is...

$$ P[T < t] = \begin{cases} \frac{\gamma(a,\frac{t}{\mu})}{\Gamma(a)}, & t \gt 0 \\ 0, & t \le 0 \end{cases} $$

Where $\mu$ is the average per trial and $a$ is the # inter-arrival times.

Answer 1

For your problem, $a = 1$ (The gamma reduces to the exponential distribution) and $\mu = 0.5$ [s/beep]... $$ P[T>5] = 1-P[T<5] = 1 - \frac{\gamma(1,2\cdot 5)}{\Gamma(1)} = 1 - (1 - e^{-10}) \\\implies P[T>5] =e^{-10} $$

Answer 2

Let $Y = Y_1 + Y_2 + Y_3$ so $a = 3$ this time... $$ P[Y=4] = \frac{4^{3 - 1}e^{-2\cdot 4}}{0.5^3\Gamma(3)} \\\implies P[Y=4] = 64e^{-8} $$

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  • $\begingroup$ The notation $P(T=t)$ to denote the PDF of a continuous random variable is misleading since the true probability $P(T=t)=0$ for any continuous random variable. I suggest using the standard density function notation $f_T(t)$. Remember the value of a density function is not a probability but a probability per unit and hence can be $>1$, yet another reason why $P(T=t)$ is not really suitable to denote the PDF. $\endgroup$ Nov 8, 2019 at 23:09

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