Questions

Procrastinators

Three students are working independently on their Stat 110 problem set. All three start at 1pm on the day the pset is due, and each takes an Exponential time with mean 6 hours to complete the homework.

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What is the earliest time when all three students will have completed the homework, on average?

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Label the students, and let XiX_i be the amount of time it takes student ii to finish the pset. Let TT be the total amount of time it takes for all three students to finish, and let T=T1+T2+T3T = T_1 + T_2 + T_3, where T1=min(X1,X2,X3)T_1 = \min(X_1, X_2,X_3), and so on. We know that T1Expo(36)T_1 \sim Expo(\frac{3}{6}), as we showed in the previous problem set, T2Expo(26)T_2 \sim Expo(\frac{2}{6}), and T3Expo(16)T_3\sim Expo(\frac{1}{6}). Therefore,

E[T]=2+3+6=11E[T] = 2 + 3 + 6 = \boxed{11}

Counting Cars

Cars pass by a certain point on a road according to a Poisson process with rate λ\lambda cars/minute. Let NtPois(λt)N_t \sim Pois(\lambda t) be the number of cars that pass by that point in the time interval [0,t][0, t], with tt measured in minutes.

A certain device is able to count cars as they pass by, but it does not record the arrival times. At time 0, the counter on the device is reset to 0. At time 3 minutes, the device is observed and it is found that exactly 1 car had passed by.

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Given this information, find the conditional CDF of when that car arrived. Also describe in words what the result says.

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Let T1T_1 be the arrival time of the first car to arrive after time 0. Unconditionally, T1Expo(λ)T_1 \sim Expo(\lambda). Given N3=1N_3 = 1, for 0t30 \leq t \leq 3, we have P(T1tN3=1)=P(Nt1N3=1)=P(Nt1,N3=1)P(N3=1)=P(Nt=1,N3=1)P(N3=1)P(T_1 \leq t | N_3 = 1) = P(N_t \geq 1 | N_3 = 1) = \frac{P(N_t \geq 1, N_3 = 1)}{P(N_3 = 1)} = \frac{P(N_t = 1, N_3 = 1)}{P(N_3 = 1)} By definition of the Poisson process, the numerator is

P(Nt=1,N3=1)=P(N[0,t]=1,N(t,3]=0)=P(N[0,t]=1)P(N(t,3]=0)=eλtλteλ(3t)=λte3λ\begin{aligned} P(N_t = 1, N_3 = 1) &= P\left({N_{[0, t]} = 1, N_{(t, 3]} = 0}\right) \\ &= P\left({N_{[0, t]} = 1}\right)P\left({N_{(t, 3]} = 0}\right)\\ &= e^{-\lambda t}\lambda t e^{-\lambda(3 - t)}\\ &= \lambda t e^{-3 \lambda} \end{aligned}

and the denominator is e3λ3λe^{-3\lambda} 3 \lambda. Hence,

P(T1tN3=1)=λte3λe3λ3λ=t3P(T_1 \leq t | N_3 = 1) = \frac{\lambda t e^{-3 \lambda}}{e^{-3\lambda} 3 \lambda} = \frac{t}{3}

for 0t30 \leq t \leq 3 (and 0 for t<0t < 0 and 1 for t>3t > 3). This says that the conditional distribution of the first arrival time, given that there was exactly one arrival in [0,3][0, 3], is Unif(0,3)Unif(0, 3).

In the late afternoon, you are counting blue cars. Each car that passes by is blue with probability $b$, independently of all other cars.

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Find the marginal PMF of the number of blue cars and number of non-blue cars that pass by the point in 10 minutes.

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Let XX and YY be the number of blue and non-blue cars that pass by in those 10 minutes respectively, and N=X+YN = X + Y. Then NPois(10λ)N \sim Pois(10\lambda) and XNBin(N,b)X|N \sim Bin(N, b). By the chicken-egg story, XX and YY are independent with

XPois(10λb)X \sim Pois(10 \lambda b)
YPois(10λ(1b))Y \sim Pois(10 \lambda (1 - b))
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Find the joint PMF of the number of blue cars and number of non-blue cars that pass by the point in 10 minutes.

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The joint PMF is the product of the marginal PMFs:

P(X=x,Y=y)=e10λb(10λb)xx!e10λ(1b)(10λ(1b))yy!P(X = x, Y = y) = \frac{e^{-10 \lambda b}(10 \lambda b)^x}{x!}\frac{e^{-10 \lambda (1 - b)}(10 \lambda (1 - b))^y}{y!}

for all nonnegative integers x,yx, y.

Normal Moments

Let X1N(μ1,σ12)X_1 \sim N(\mu_1, \sigma_1^2) and X2N(μ2,σ22)X_2 \sim N(\mu_2, \sigma_2^2).

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Use the MGF to show that for any values of a,b̸=0a, b \not= 0, Y=aX1+bX2Y = a X_1 + b X_2 is also Normal. (You may use the fact that the MGF of a N(μ,σ2)N(\mu, \sigma^2) distribution is eμt+12σ2t2e^{\mu t + \frac{1}{2}\sigma^2 t^2}.)

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We know that the MGF for the sum of 2 independent random variables is the product of their individual MGFs. We also know that the MGF of a scalar cc times a random variable is equal to that random variable's MGF with ctct subsituted for tt.

MY(t)=exp(μ1(at)+12σ12(at)2)exp(μ2(bt)+12σ22(bt)2)=exp(μ1(at)+12σ12(at)2+μ2(bt)+12σ22(bt)2)=exp((aμ1+bμ2)t+(aσ1)2+(bσ2)22t2)\begin{aligned} M_Y(t) &= \exp \left(\mu_1 (at) + \frac{1}{2} \sigma_1^2 (at)^2\right) \cdot \exp\left(\mu_2 (bt) + \frac{1}{2} \sigma_2^2 (bt)^2\right) \\ &= \exp\left( \mu_1 (at) + \frac{1}{2} \sigma_1^2 (at)^2 + \mu_2 (bt) + \frac{1}{2} \sigma_2^2 (bt)^2\right) \\ &= \exp\left( (a\mu_1 + b \mu_2)t + {\frac{(a \sigma_1)^2 + (b \sigma_2)^2}{2} } t^2\right) \end{aligned}

This is the N(aμ1+bμ2,a2σ12+b2σ22)N\left({a \mu_1 + b \mu_2, a^2 \sigma_1^2 + b^2 \sigma_2^2}\right) MGF.

Let ZN(0,1)Z \sim N(0, 1) and Y=eZY = e^Z. Then YY has a LogNormal distribution (because its log is Normal!).

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Find all the moments of YY. That is, find E(Yn)E(Y^n) for all nn.

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Using exponent rules,

E(Yn)=E((eZ)n)=E(enZ)E\left({Y^n}\right) = E\left({\left({e^Z}\right)^n}\right) = E\left({e^{nZ}}\right)

We may be tempted to use LOTUS at this point, but it is easier to recognize that E(enZ)E\left({e^{nZ}}\right) is the MGF of ZZ, evaluated at t=nt = n. In other words, E(enZ)=MZ(n)E\left({e^{nZ}}\right) = M_Z(n). Since MZ(t)=et2/2M_Z(t) = e^{t^2/2}, we conclude that MZ(n)=en2/2M_Z(n) = e^{n^2/2}.

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Show the MGF of YY does not exist, by arguing that the integral E(etY)E\left({e^{tY}}\right) diverges for t>0t > 0.

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The key is to remember that the MGF is an expectation, which requires LOTUS and integration:

E(etY)=E(eteZ)=etez12πez2/2dz=12πetezz2/2dzE\left({e^{tY}}\right) = E\left({e^{te^Z}}\right) = \int_{-\infty}^\infty e^{te^z} \frac{1}{\sqrt{2 \pi}} e^{-z^2/2} \textrm{d} z = \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi}} e^{te^z -z^2/2} \textrm{d} z

Suppose t>0t > 0. Then as zz grows larger, tezte^z will grow at a faster rate than z2/2z^2 / 2. Hence, tezz2/2te^z - z^2 / 2 will explode. Therefore, the integral diverges for all t>0t > 0. Since E(etY)E\left({e^{tY}}\right) is not finite on an open interval around 0, the MGF of YY does not exist.