Questions

Birthdays

Use Poisson approximations to investigate the following types of coincidences. The usual assumptions of the birthday problem apply, such as that there are 365 days in a year, with all days equally likely.

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How many people are needed to have a 50% chance that at least one of them has the same birthday as you?

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Let kk be the number of people there are other than you. Create an indicator variable IiI_i for each of the kk people as to whether they have the same birthday as you. Then, P(Ii=1)=1365P(I_i = 1) = \frac{1}{365} and thus E[i=1kIi]=k365E\left[ \sum_{i = 1}^{k} I_i \right] = \frac{k}{365}. Therefore, we can model this as a Pois(k365)\text{Pois} \left( \frac{k}{365} \right) and so we just need to calculate

1ek/365=0.51 - e^{-k/365} = 0.5

It turns out that k253k \approx 253.

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How many people are needed to have a 50% chance that there are two people who not only were born on the same day, but also were born at the same hour (e.g., two people born between 2 pm and 3 pm are considered to have been born at the same hour).

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This is the birthday problem but with 36524365 \cdot 24 types instead of just 365. Creating an indicator r.v. for whether each pair of kk people have the same birthday, we get that the number of pairs of people with the same birthday is distributed approximately Pois((k2)36524)\text{Pois} \left( \frac{{k \choose 2}}{365 \cdot 24} \right) and thus the probability of at least people having the same birthday is approximately:

1e(k2)365241 - e^{-\frac{{k \choose 2}}{365 \cdot 24}}

Setting it equal to 12\frac{1}{2} gives us k=111k = 111.

Uniform

Let $U \sim Unif(-1,1).$ Note that the PDF of $U$ is $f(x) = \frac{1}{2}$.

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Compute E(U)E(U)

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E(U)=0E(U) = 0 because the distribution is symmetric about 0.

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Compute Var(U)Var(U)

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We need to calculate E(U2)E(U^2) for the variance, so we have:

E(U2)=11u212du=[16u3]11=13E(U^2) = \int_{-1}^{1} u^2 \cdot \frac{1}{2} d u = \left[\frac{1}{6} u^3\right]_{-1}^{1} = \frac{1}{3}

Therefore, Var(U)=E(U2)E(U)2=13Var\left({U}\right) = E(U^2) - E(U)^2 = \boxed{\frac{1}{3}}

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Compute E(U4)E(U^4)

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We use LOTUS (Law of the Unconscious Statistician) as before.

E(U4)=11u412du=[110u5]11=15E(U^4) = \int_{-1}^{1} u^4 \cdot \frac{1}{2} d u = \left[\frac{1}{10} u^5\right]_{-1}^{1} = \boxed{\frac{1}{5}}
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Find the CDF of U2U^2.

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We can describe the CDF of U2U^2 using the CDF of UU.

P(U2<k)=P(k<U<k)=2k2=kP(U^2 < k) = P(-\sqrt{k} < U < \sqrt{k}) = \frac{2\sqrt{k}}{2} = \boxed{\sqrt{k}}
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Find the PDF of U2U^2.

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We take the derivative of the CDF of U2U^2 to obtain the PDF UU.

ddkP(U2<k)=12k\frac{d}{dk} P(U^2 < k) = \boxed{\frac{1}{2 \sqrt{k}}}
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Is U2U^2 a uniform distribution?

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The PDF of U2U^2 is 12k\frac{1}{2 \sqrt{k}}. This is not a uniform distribution. This also shows that UkU^k is not uniform anymore for any k>1k > 1.

Normal

Let ZN(0,1)Z \sim N(0,1) with CDF Φ\Phi. The PDF of Z2Z^2 is the function given by:

g(w)=12πwew/2g(w) = \frac{1}{\sqrt{2\pi w}} e^{-w/2}

with a support of w0w \geq 0.

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Find expressions for E(Z4)E(Z^4) as integrals in two different ways one based on the PDF of ZZ and the other based on the PDF of Z2Z^2.

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Let W=Z2,W=Z^{2}, so W2=Z4.W^{2}=Z^{4} . By LOTUS

E(Z4)=z4φ(z)dz=0w2g(w)dwE\left(Z^{4}\right)=\int_{-\infty}^{\infty} z^{4} \varphi(z) d z=\int_{0}^{\infty} w^{2} g(w) d w

where φ(z)=12πez2/2\varphi(z)=\frac{1}{\sqrt{2 \pi}} e^{-z^{2} / 2} is the PDF of Z,Z, and gg is as above.

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Find E(Z2+Z+Φ(Z))E(Z^2 + Z + \Phi(Z)).

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By linearity, this is E(Z2)+E(Z)+E(Φ(Z))E\left(Z^{2}\right)+E(Z)+E(\Phi(Z)). The second term is 0 and the first term is 1 since E(Z)=0,Var(Z)=1.E(Z)=0, \operatorname{Var}(Z)=1 . The third term is 1/21 / 2 since by universality of the Uniform, Φ(Z)Unif(0,1)\Phi(Z) \sim \operatorname{Unif}(0,1). Thus, the value is 3/23 / 2.

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Find the CDF of Z2Z^2 in terms of Φ\Phi; do not find the PDF of gg.

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For w0,w \leq 0, the CDF of Z2Z^{2} is 0. For w>0,w>0, the CDF of Z2Z^{2} is

P(Z2w)=P(wZw)=Φ(w)Φ(w)=2Φ(w)1P\left(Z^{2} \leq w\right)=P(-\sqrt{w} \leq Z \leq \sqrt{w})=\Phi(\sqrt{w})-\Phi(-\sqrt{w})=2 \Phi(\sqrt{w})-1