STAT110: Fall 2020
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  • Birthdays
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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.

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

Let kkk be the number of people there are other than you. Create an indicator variable IiI_iIi​ for each of the kkk people as to whether they have the same birthday as you. Then, P(Ii=1)=1365P(I_i = 1) = \frac{1}{365}P(Ii​=1)=3651​ and thus E[∑i=1kIi]=k365E\left[ \sum_{i = 1}^{k} I_i \right] = \frac{k}{365}E[∑i=1k​Ii​]=365k​. Therefore, we can model this as a Pois(k365)\text{Pois} \left( \frac{k}{365} \right)Pois(365k​) and so we just need to calculate

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

It turns out that k≈253k \approx 253k≈253.

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).

This is the birthday problem but with types instead of just 365. Creating an indicator r.v. for whether each pair of people have the same birthday, we get that the number of pairs of people with the same birthday is distributed approximately and thus the probability of at least people having the same birthday is approximately:

Setting it equal to gives us .

Uniform

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

Compute E(U)E(U)E(U)

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

Compute Var(U)Var(U)Var(U)

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

E(U2)=∫−11u2⋅12du=[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}E(U2)=∫−11​u2⋅21​du=[61​u3]−11​=31​

Therefore, Var(U)=E(U2)−E(U)2=13Var\left({U}\right) = E(U^2) - E(U)^2 = \boxed{\frac{1}{3}}Var(U)=E(U2)−E(U)2=31​​

Compute E(U4)E(U^4)E(U4)

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

E(U4)=∫−11u4⋅12du=[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}}E(U4)=∫−11​u4⋅21​du=[101​u5]−11​=51​​

Find the CDF of U2U^2U2.

We can describe the CDF of U2U^2U2 using the CDF of UUU.

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}}P(U2<k)=P(−k​<U<k​)=22k​​=k​​

Find the PDF of U2U^2U2.

We take the derivative of the CDF of U2U^2U2 to obtain the PDF UUU.

ddkP(U2<k)=12k\frac{d}{dk} P(U^2 < k) = \boxed{\frac{1}{2 \sqrt{k}}}dkd​P(U2<k)=2k​1​​

Is U2U^2U2 a uniform distribution?

The PDF of U2U^2U2 is 12k\frac{1}{2 \sqrt{k}}2k​1​. This is not a uniform distribution. This also shows that UkU^kUk is not uniform anymore for any k>1k > 1k>1.

Normal

Let Z∼N(0,1)Z \sim N(0,1)Z∼N(0,1) with CDF Φ\PhiΦ. The PDF of Z2Z^2Z2 is the function given by:

g(w)=12πwe−w/2g(w) = \frac{1}{\sqrt{2\pi w}} e^{-w/2}g(w)=2πw​1​e−w/2

with a support of w≥0w \geq 0w≥0.

Find expressions for as integrals in two different ways one based on the PDF of and the other based on the PDF of .

Let so By LOTUS

where is the PDF of and is as above.

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

By linearity, this is E(Z2)+E(Z)+E(Φ(Z))E\left(Z^{2}\right)+E(Z)+E(\Phi(Z))E(Z2)+E(Z)+E(Φ(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 .E(Z)=0,Var(Z)=1. The third term is since by universality of the Uniform, . Thus, the value is .

Find the CDF of Z2Z^2Z2 in terms of Φ\PhiΦ; do not find the PDF of ggg.

For the CDF of is 0. For the CDF of is

365⋅24365 \cdot 24365⋅24
kkk
Pois((k2)365⋅24)\text{Pois} \left( \frac{{k \choose 2}}{365 \cdot 24} \right)Pois(365⋅24(2k​)​)
1−e−(k2)365⋅241 - e^{-\frac{{k \choose 2}}{365 \cdot 24}}1−e−365⋅24(2k​)​
12\frac{1}{2}21​
k=111k = 111k=111
E(Z4)E(Z^4)E(Z4)
ZZZ
Z2Z^2Z2
W=Z2,W=Z^{2},W=Z2,
W2=Z4.W^{2}=Z^{4} .W2=Z4.
E(Z4)=∫−∞∞z4φ(z)dz=∫0∞w2g(w)dwE\left(Z^{4}\right)=\int_{-\infty}^{\infty} z^{4} \varphi(z) d z=\int_{0}^{\infty} w^{2} g(w) d wE(Z4)=∫−∞∞​z4φ(z)dz=∫0∞​w2g(w)dw
φ(z)=12πe−z2/2\varphi(z)=\frac{1}{\sqrt{2 \pi}} e^{-z^{2} / 2}φ(z)=2π​1​e−z2/2
Z,Z,Z,
ggg
1/21 / 21/2
Φ(Z)∼Unif⁡(0,1)\Phi(Z) \sim \operatorname{Unif}(0,1)Φ(Z)∼Unif(0,1)
3/23 / 23/2
w≤0,w \leq 0,w≤0,
Z2Z^{2}Z2
w>0,w>0,w>0,
Z2Z^{2}Z2
P(Z2≤w)=P(−w≤Z≤w)=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})-1P(Z2≤w)=P(−w​≤Z≤w​)=Φ(w​)−Φ(−w​)=2Φ(w​)−1