> For the complete documentation index, see [llms.txt](https://justinthezhu.gitbook.io/stat110/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://justinthezhu.gitbook.io/stat110/untitled-4/questions.md). # 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. {% tabs %} {% tab title="Q" %} How many people are needed to have a 50% chance that at least one of them has the same birthday as *you*? {% endtab %} {% tab title="A" %} 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$$. {% endtab %} {% endtabs %} {% tabs %} {% tab title="Q" %} How many people are needed to have a 50% chance that there are two people who not only were born on the same day, but also were born at the same hour (e.g., two people born between 2 pm and 3 pm are considered to have been born at the same hour). {% endtab %} {% tab title="A" %} This is the birthday problem but with $$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$$. {% endtab %} {% endtabs %} ## Uniform Let $U \sim Unif(-1,1).$ Note that the PDF of $U$ is $f(x) = \frac{1}{2}$. {% tabs %} {% tab title="Q" %} Compute $$E(U)$$ {% endtab %} {% tab title="A" %} $$E(U) = 0$$ because the distribution is symmetric about 0. {% endtab %} {% endtabs %} {% tabs %} {% tab title="Q" %} Compute $$Var(U)$$ {% endtab %} {% tab title="A" %} We need to calculate $$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}}$$ {% endtab %} {% endtabs %} {% tabs %} {% tab title="Q" %} Compute $$E(U^4)$$ {% endtab %} {% tab title="A" %} We use LOTUS (Law of the Unconscious Statistician) as before. $$ 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}} $$ {% endtab %} {% endtabs %} {% tabs %} {% tab title="Q" %} Find the CDF of $$U^2$$. {% endtab %} {% tab title="A" %} We can describe the CDF of $$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}} $$ {% endtab %} {% endtabs %} {% tabs %} {% tab title="Q" %} Find the PDF of $$U^2$$. {% endtab %} {% tab title="A" %} We take the derivative of the CDF of $$U^2$$ to obtain the PDF $$U$$. $$ \frac{d}{dk} P(U^2 < k) = \boxed{\frac{1}{2 \sqrt{k}}} $$ {% endtab %} {% endtabs %} {% tabs %} {% tab title="Q" %} Is $$U^2$$ a uniform distribution? {% endtab %} {% tab title="A" %} The PDF of $$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$$. {% endtab %} {% endtabs %} ## Normal Let $$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$$. {% tabs %} {% tab title="Q" %} Find expressions for $$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$$. {% endtab %} {% tab title="A" %} Let $$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$$ is as above. {% endtab %} {% endtabs %} {% tabs %} {% tab title="Q" %} Find $$E(Z^2 + Z + \Phi(Z))$$. {% endtab %} {% tab title="A" %} By linearity, this is $$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$$. {% endtab %} {% endtabs %} {% tabs %} {% tab title="Q" %} Find the CDF of $$Z^2$$ in terms of $$\Phi$$; do not find the PDF of $$g$$. {% endtab %} {% tab title="A" %} For $$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 $$ {% endtab %} {% endtabs %} --- # Agent Instructions This documentation is published with GitBook. GitBook is the documentation platform designed so that both humans and AI agents can read, navigate, and reason over technical content effectively. 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