Questions

Housing Day

Suppose Harvard College is conducting its housing lottery. For simplicity's sake, we'll say that there are 1200 Freshmen that will be randomly assigned to 12 houses. Let X1,X2,,X12X_1, X_2, \ldots, X_{12} count how many students are place in Pforzheimer (X1X_1), all the way to Eliot (X12X_{12}) (organized by best house to worst).

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Are X1X_1 and X2X_2 independent?

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No they are not. Since the number of Freshmen is constrained to 1200, knowing that a lot of people got into one house decreases the number of people that could be in the remaining houses.

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What is the joint distribution of X1,X2,,X12X_1, X_2, \ldots, X_{12}?

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By the story of the Multinomial distribution, (X1,X2,,X12)Mult12(1200,(1/12,,1/12))(X_1, X_2, \ldots, X_{12}) \sim Mult_{12}\left({1200, \left({1/12, \dots, 1/12}\right)}\right)

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What is the marginal distribution of X1X_1, the number of students who are placed into Pforzheimer House, and the joint distribution of X1X_1 and 1200X11200 - X_1?

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In this case, we can group together bins that are not in Pforzheimer House together. We have X1Bin(1200,1/12)X_1 \sim Bin\left({1200, 1/12}\right) (X1,1200X1)Mult2(1200,(1/12,11/12))(X_1, 1200 - X_1) \sim Mult_2\left({1200, (1/12, 11/12)}\right)

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What is the conditional distribution of X1X_1 given X10+X11+X12=450X_{10} + X_{11} + X_{12} = 450?

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X1X10+X11+X12=450Bin(750,1/9)X_1 | X_{10} + X_{11} + X_{12} = 450 \sim Bin\left({750, 1/9}\right)

Jelly Beans

I have a jar of 30 jellybeans: 10 red, 8 green, 12 blue. I draw a sample of 12 jellybeans without replacement. Let XX be the number of red jellybeans in the sample, YY the number of green jellybeans.

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Find Cov(X,Y)Cov(X, Y).

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Let X=I1++I12X = I_1 + \ldots + I_{12}, and Y=J1++J12Y = J_1 + \ldots + J_{12}, where

Ii={1if ith jellybean in sample is red0otherwiseJi={1if ith jellybean in sample is green0otherwise\begin{aligned} I_i &= \begin{cases} 1 & \textrm{if $$i$$th jellybean in sample is red} \\ 0 & \textrm{otherwise} \end{cases} \\ J_i &= \begin{cases} 1 & \textrm{if $$i$$th jellybean in sample is green} \\ 0 & \textrm{otherwise} \end{cases}\end{aligned}

We can now solve using indicator variables and the fundamental bridge.

Cov(I1,J1)=E(I1J1)E(I1)E(J1)=0(1030)(830)Cov(I1,J2)=E(I1J2)E(I1)E(J2)=(1030)(829)(1030)(830)Cov(X,Y)=i=112Cov(Ii,Ji)+2i<jCov(Ii,Jj)=i=112Cov(I1,J1)+2(122)Cov(I1,J2)=12Cov(I1,J1)+1211Cov(I1,J2)=96145\begin{aligned} Cov\left({I_1, J_1}\right) &= E\left({I_1 J_1}\right) - E\left({I_1}\right) E\left({J_1}\right) \\ &= 0 - \left({\frac{10}{30}}\right)\left({\frac{8}{30}}\right) \\ Cov\left({I_1, J_2}\right) &= E\left({I_1 J_2}\right) - E\left({I_1}\right) E\left({J_2}\right) \\ &= \left({\frac{10}{30}}\right)\left({\frac{8}{29}}\right) - \left({\frac{10}{30}}\right)\left({\frac{8}{30}}\right) \\ Cov\left({X, Y}\right) &= \sum_{i=1}^{12} Cov\left({I_i, J_i}\right) + 2 \sum_{i < j} Cov\left({I_i, J_j}\right) \\ &= \sum_{i=1}^{12} Cov\left({I_1, J_1}\right) + 2 \binom{12}{2} Cov\left({I_1, J_2}\right) \\ &= 12 \cdot Cov\left({I_1, J_1}\right) + 12 \cdot 11 \cdot Cov\left({I_1, J_2}\right) \\ &= - \frac{96}{145}\end{aligned}

It's good to do a little sanity check at the end: it makes sense that the covariance is negative. If the sample contains a lot of red jellybeans, the sample probably has fewer green jellybeans. Another way to solve this is to create an indicator for each red jellybean and each green jellybean in the jar, where the indicator equals 1 if the jellybean is in the sample and 0 otherwise.

Stat Courses

Let XX be the number of statistics majors in a certain college in the class of 2030, viewed as an r.v. Each statistics major chooses between two tracks: a general track in statistical principles, and a track in quant finance. Suppose that each statistics major chooses randomly which of these two tracks to follow, independently, with probability pp of choosing the general track. Let YY be the number of statistics majors who choose the general track, and ZZ be the number of statistics majors who choose the quantitative finance track.

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Suppose that XPois(λ)X \sim \text{Pois}(\lambda). Find the correlation between XX and YY.

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By the chicken-egg story, we know that YY and ZZ are independent Poisson random variables, with rate parameters λp\lambda p and λq\lambda q, respectively. We must first find the covariance between XX and YY.

Cov(X,Y)=Cov(Y+Z,Y)=Var(Y)+Cov(Y,Z)=λpCov(X,Y) = Cov(Y+Z, Y) = Var(Y) + Cov(Y,Z) = \lambda p

We now plug this into the equation for correlation:

Corr(X,Y)=Cov(X,Y)Var(X)Var(Y)=λpλλp=pCorr(X,Y) =\frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}} = \frac{\lambda p}{\sqrt{\lambda \lambda p}} = \sqrt{p}
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Let nn be the size of the Class of 2030, where nn is a known constant. For this part and the next, instead of assuming that XX is Poisson, assume that each of the nn students chooses to be a statistics major with probability rr, independently. Find the joint distributions of YY, ZZ, and the number of non-statistics majors, and their marginal distributions.

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Under this new model, we have that XBin(n,r)X\sim \text{Bin}(n,r). By the multiplication rule, we have that the probability of a student becoming a general Statistician is rprp, a Goldman-Sachs Statistician is rqrq, and a non-Statstician (lame) is 1r1-r. Therefore, we can apply the story of the Multinomial here: (Y,Z,nX)Mult3(n,(rp,rq,1r))(Y, Z, n-X) \sim Mult_3(n, (rp, rq, 1-r))

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Continuing as in the previous part, find the correlation between XX and YY.

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We use the fact that covariance of the marginal distributions in a multinomial is given by npipj-np_i p_j.

Cov(X,Y)=Cov(Y+Z,Y)=Var(Y)+Cov(Z,Y)=nrp(1rp)n(rp)(rq)=npr(1r)\begin{aligned} Cov(X,Y) &= Cov(Y+Z, Y)\\ &= Var(Y) + Cov(Z,Y) \\ &= nrp(1-rp) - n(rp)(rq)\\ &=npr(1-r) \end{aligned}