The cumulative distribution function of T

cumulative-distribution-functionsexponential distributionprobabilityprobability distributions

Define 3 independent random variables:

  • X the time until your friend Alex calls, with an average waiting time for his call 1/λ (λ > 0)
  • Y the time until your friend Jessica calls, with an average waiting time for her call 1/μ (μ > 0)
  • Z the time until your friend Mick calls, with an average waiting time for her call 1/ν (ν > 0)

  1. Let T be the time before any of those calls you. What is the cumulative distribution function of T?

  2. Derive the p.m.f. of $I$:
    \begin{equation}
    I=
    \begin{cases}
    0 & \text{if }X<min\{Y,Z\}\\
    1 & \text{if }Y<min\{X,Z\}\\
    2 & \text{if }Z<min\{X,Y\}\\
    \end{cases}
    \end{equation}

My Attempt

I recognize each r.v. as exponential distributions, so we have
$$X\sim exp(1/\lambda)\implies f_X(x)=\frac{1}{\lambda}e^{-\frac{1}{\lambda}x}\implies F_X(x)=1-e^{-\frac{t}{\lambda}}$$
$$Y\sim exp(1/\mu)\implies f_Y(y)=\frac{1}{\mu}e^{-\frac{1}{\mu}y}\implies F_Y(y)=1-e^{-\frac{t}{\mu}}$$
$$Z\sim exp(1/v)\implies f_Z(z)=\frac{1}{v}e^{-\frac{1}{v}z}\implies F_Z(z)=1-e^{-\frac{t}{v}}$$
Then $T=min(X,Y,Z)$ so our c.d.f. is
$$F_T(t)=F_X(x)\cdot F_Y(y)\cdot F_Z(z)=(1-e^{-\frac{t}{\lambda}})(1-e^{-\frac{t}{\mu}})(1-e^{-\frac{t}{v}})$$
My question is what are the ranges for this c.d.f.? When is the probability $0$? When is it $1$?

I'm not sure what the second question is asking because isn't what's provided already a p.m.f.? What are we supposed to derive?

Best Answer

We are given the mean so the rate is the inverse of the provided quantities.

$$X\sim \text{Exponential}(\lambda)\\ Y\sim \text{Exponential}(\mu)\\ Z\sim \text{Exponential}(\nu)$$

One way to find the cdf of $T$ is to consider:

$$\begin{split}Pr(T>t)&=Pr(\min\{X,Y,Z\}>t)\\ &=Pr(X>t)Pr(Y>t)Pr(Z>t)\\ &=e^{-\lambda t}e^{-\mu t}e^{-\nu t}\end{split}$$

Thus the cdf is $$1-e^{-(\lambda+\mu+\nu)t}, t>0$$

For part 2, the task is to find the probabilities. Note that $\min\{Y,Z\}\sim\text{Exponential}(\mu+\nu)$ by the previous problem. Define $W=\min\{Y,Z\}$.

$$\begin{split}Pr(X<W)&=\int_0^\infty Pr(X<W|W)f(W)dW\\ &=\int_0^\infty \left(1-e^{-\lambda w}\right)(\mu+\nu) e^{-(\mu+\nu)w}dw\\ &=1-\frac{\mu+\nu}{\lambda+\mu+\nu} \end{split}$$

By same pattern we can fil out

$$Pr(I=0)=1-\frac{\mu+\nu}{\lambda+\mu+\nu}\\ Pr(I=1)=1-\frac{\lambda+\nu}{\lambda+\mu+\nu}\\ Pr(I=2)=1-\frac{\lambda+\mu}{\lambda+\mu+\nu}$$

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