Questions

Chicken and Egg

Recall in the Chicken-Egg problem that we have a chicken that lays eggs, where the number of eggs NN follows the distribution NPois(λ)N\sim Pois(\lambda). Given that the chicken lays NN eggs, the number of eggs that hatch XX follows the distribution XNBin(N,p)X|N \sim Bin(N,p).

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Find E(X)E(X)

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Use Adam's Law!

E(X)=E[E(XN)]=E[Np]=pE[N]=λp\begin{aligned} E(X) &= E[E(X|N)] = E[Np] = p E[N] =\lambda p \end{aligned}
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Find Var(X)Var(X)

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Use Eve's Law!

Var(X)=E[Var(XN)]+Var[E(XN)]=E[Np(1p)]+Var[Np]=λp(1p)+p2λ=λp\begin{aligned} Var(X) &= E[Var(X|N)] + Var[E(X|N)] \\ &= E[Np(1-p)] + Var[Np] \\ &= \lambda p(1-p) + p^2 \lambda \\ &= \lambda p \end{aligned}

Consumerism

When customers enter a particular store, each makes a purchase with probability pp, independently. Given that a customer makes a purchase, the amount spent has mean μ\mu (in dollars) and variance σ2\sigma^2.

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Find the mean of how much a random customer spends (note that the customer may spend nothing).

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Let XX be the amount a random customer spends at the store, and let II be the indicator that a random customer makes a purchase. We then apply the Law of Total Expectation:

E[X]=E[XI=0]P(I=0)+E[XI=1]P(I=1)=μp\begin{aligned} E[X] &= E[X | I= 0]P(I=0) \\&+ E[X | I=1]P(I=1)\\ &= \mu p \end{aligned}
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Find the variance of how much a random customer spends (note that the customer may spend nothing).

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Let XX be the amount a random customer spends at the store, and let II be the indicator that a random customer makes a purchase. We then apply the Law of Total Expectation:

E[X2]=E[X2I=0]P(I=0)+E[X2I=1]P(I=1)=E[X2I=1]P(I=1)=(μ2+σ2)pVar[X]=E[X2]E[X]2=σ2p+μ2p(1p)\begin{aligned} E[X^2] &= E[X^2 | I=0]P(I=0) \\ &+ E[X^2 | I=1]P(I=1) \\ &= E[X^2 | I=1]P(I=1) \\ &= (\mu^2 + \sigma^2)p \\ Var[X] &= E[X^2] - E[X]^2 \\ &= \sigma^2 p +\mu^2 p (1-p) \end{aligned}

Trapped Miners

A miner is trapped in a mine containing 3 doors. The first door leads to a tunnel that will take him to safety after 3 minutes. The other two doors lead to tunnels that will return him to the mine after 5 and 7 minutes of travel each. The miner is equally likely to choose any of the doors at any time.

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What is the expected amount of time until he reaches safety?

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Let WW be the total amount of time taken until he reaches safety, and D1D_1, D2D_2, and D3D_3 be the events of going through doors 1, 2, and 3 respectively.

Using LoTE, we have

E(W)=E(WD1)P(D1)+E(WD2)P(D2)+E(WD3)P(D3)=313+(5+E(W))13+(7+E(W))1313E(W)=5E(W)=15\begin{aligned} E(W) &= E(W|D_1)P(D_1) + E(W|D_2)P(D_2) + E(W|D_3)P(D_3) \\ &= 3 \frac{1}{3} + (5 + E(W)) \frac{1}{3} + (7 + E(W)) \frac{1}{3} \\ \implies\frac{1}{3} E(W) &= 5 \\ E(W) &= 15\end{aligned}