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

$1 - e^{-k/365} = 0.5$

It turns out that $k \approx 253$.

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

$1 - e^{-\frac{{k \choose 2}}{365 \cdot 24}}$

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

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

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Compute $E(U)$

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

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Compute $Var(U)$

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

$E(U^2) = \int_{-1}^{1} u^2 \cdot \frac{1}{2} d u = \left[\frac{1}{6} u^3\right]_{-1}^{1} = \frac{1}{3}$

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

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Compute $E(U^4)$

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

$E(U^4) = \int_{-1}^{1} u^4 \cdot \frac{1}{2} d u = \left[\frac{1}{10} u^5\right]_{-1}^{1} = \boxed{\frac{1}{5}}$

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Find the CDF of $U^2$.

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We can describe the CDF of $U^2$ using the CDF of $U$.

$P(U^2 < k) = P(-\sqrt{k} < U < \sqrt{k}) = \frac{2\sqrt{k}}{2} = \boxed{\sqrt{k}}$

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Find the PDF of $U^2$.

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We take the derivative of the CDF of $U^2$ to obtain the PDF $U$.

$\frac{d}{dk} P(U^2 < k) = \boxed{\frac{1}{2 \sqrt{k}}}$

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Is $U^2$ a uniform distribution?

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

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

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

with a support of $w \geq 0$.

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

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Let $W=Z^{2},$ so $W^{2}=Z^{4} .$ By LOTUS

$E\left(Z^{4}\right)=\int_{-\infty}^{\infty} z^{4} \varphi(z) d z=\int_{0}^{\infty} w^{2} g(w) d w$

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

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Find $E(Z^2 + Z + \Phi(Z))$.

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

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Find the CDF of $Z^2$ in terms of $\Phi$; do not find the PDF of $g$.

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

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