Problems

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Solution:

1. \(\,\) \begin{align*} && \mathbb{E}(Z| a < Z < b) &= \mathbb{E}(Z\mathbb{1}_{(a,b)}) /\mathbb{E}(\mathbb{1}_{(a,b)}) \\ &&&= \int_a^b z \phi(z) \d z \Big / (\Phi(b) - \Phi(a)) \\ &&&= \frac{\int_a^b \frac{1}{\sqrt{2 \pi}}z e^{-\frac12 z^2} \d z}{\Phi(b) - \Phi(a)} \\ &&&= \frac{\frac1{\sqrt{2\pi}} \left [-e^{-\frac12 z^2} \right]_a^b}{\Phi(b) - \Phi(a)} \\ &&&= \frac{\frac1{\sqrt{2\pi}} \left (e^{-\frac12 a^2}-e^{-\frac12 b^2} \right)}{\Phi(b) - \Phi(a)} \\ \end{align*}
2. \(\,\) \begin{align*} && \mathbb{E}(X |X > 0) &= \mathbb{E}(\mu + \sigma Z | \mu + \sigma Z > 0) \\ &&&= \mathbb{E}(\mu + \sigma Z | Z > -\tfrac{\mu}{\sigma}) \\ &&&= \mathbb{E}(\mu| Z > -\tfrac{\mu}{\sigma})+ \sigma \mathbb{E}(Z | Z > -\tfrac{\mu}{\sigma})\\ &&&= \mu+ \sigma \mathbb{E}(Z | Z > -\tfrac{\mu}{\sigma})\\ \end{align*} Hence \begin{align*} &&\mathbb{E}(|X|) &= \mathbb{E}(X | X > 0)\mathbb{P}(X > 0) - \mathbb{E}(X | X < 0)\mathbb{P}(X < 0) \\ &&&=\left ( \mu+ \sigma \mathbb{E}(Z | Z > -\mu /\sigma)\right)(1-\Phi(-\mu/\sigma)) - \left ( \mu+ \sigma \mathbb{E}(Z | Z < -\mu /\sigma)\right)\Phi(-\mu/\sigma) \\ &&&= \mu(1 - 2\Phi(-\mu/\sigma)) + \sigma \frac{e^{-\frac12\mu^2/\sigma^2}}{\sqrt{2\pi}(1-\Phi(-\mu/\sigma))}(1-\Phi(-\mu/\sigma)) + \sigma \frac{e^{-\frac12\mu^2/\sigma^2}}{\sqrt{2 \pi} \Phi(-\mu/\sigma)} \Phi(-\mu/\sigma) \\ &&&= \mu(1 - 2\Phi(-\mu/\sigma)) + \sigma \sqrt{\frac{2}{\pi}} \exp(-\tfrac12 \mu^2/\sigma^2) \end{align*} Finally, \begin{align*} && \textrm{Var}(|X|) &= \mathbb{E}(|X|^2) - [\mathbb{E}(|X|)]^2 \\ &&&= \mu^2 + \sigma^2 - m^2 \end{align*}

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Solution:

1. \(X_1 \sim B(k, \tfrac12)\) so \(\E[X_1] = \frac{k}{2}\) Note that \(X_2 | X_1 = x_1 \sim B(x_1, \tfrac12)\) so \(\E[X_2 | X_1 = x_1) = \frac{x_1}{2}\) or \(\E[X_2 | X_1] = \frac12 X_1\). Therefore by the tower law, \(\E[\E[X_2|X_1]] = \E[\frac12 X_1] = \frac14k\) Notice also that \(\E[X_n] = \frac1{2^n} k\) and so \begin{align*} && \sum_{i=1}^\infty \E[X_i] &= \sum_{i=1}^{\infty} \frac1{2^i} k \\ &&&= \frac{\frac12 k}{1-\frac12} = k \end{align*}
2. Note that \(Y_1 \sim Geo(\tfrac12)-1\) which has generating function \(\E[t^{Y_1}] = \E[t^{G-1}] = \frac{\frac12 t}{1-(1-\frac12)t}\frac1{t} = \frac{\frac12}{1-\frac12t}\). Notice that \begin{align*} && \E \left [ t^Y \right] &= \E \left [ t^{\sum_{i=1}^kY_i} \right] \\ &&&= \prod_{i=1}^k \E[t^{Y_i}] \\ &&&= \frac{1}{(2-t)^k} \end{align*} Therefore \(\E[Y] = G'(1) = k(2-1)^{-(k+1)} = k\) \(\E[Y^2] = (tG'(t))'|_{t=1} = k(k+1)(2-1)^{-(k+2)}+k(2-1)^{-(k+1)} = k^2+2k\) so \(\var[Y] = k^2+2k - k^2 =2 k\). Finally \(\mathbb{P}(Y=r) = \binom{k+r-1}{k} \frac{1}{2^{r+k}}\)

[Note: this second distribution is a negative binomial distribution]

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Solution: \begin{align*} && \E[N] &= \sum_{i=1}^{2n-1} \frac{i}{2n-1} \\ &&&= \frac{2n(2n-1)}{2(2n-1)} = n\\ && \E[N^2] &= \sum_{i=1}^{2n-1} \frac{i^2}{2n-1} \\ &&&= \frac{(2n-1)(2n)(4n-1)}{6(2n-1)} \\ &&&= \frac{n(4n-1)}{3} \\ && \var[N] &= \frac{n(4n-1)}{3} - n^2 \\ &&&= \frac{n^2-n}{3} \end{align*} \begin{align*} && \E[Y] &= \E \left [ \E \left [ \sum_{i=1}^N X_i | N = k\right] \right]\\ &&&= \E \left[ N\mu \right] = n\mu \\ \\ && \mathrm{Cov}(Y,N) &= \mathbb{E}[XY] - \E[X]\E[Y] \\ &&&= \E \left [ \E \left [N \sum_{i=1}^N X_i | N = k\right] \right] - n^2 \mu \\ &&&= \E[N^2\mu] - n^2 \mu \\ &&&= \left ( \frac{n^2(4n-1)}{3} - n^2 \right) \mu \\ &&&= \frac{n^2-n}{3}\mu \\ \\ && \E[Y^2] &= \E \left [ \E \left [ \left ( \sum_{i=1}^N X_i \right) ^2\right ] \right] \\ &&&= \E \left [ \E \left [ \sum_{i=1}^N X_i ^2 + 2\sum_{i,j} X_iX_j\right ] \right] \\ &&&= \E \left [ \sum_{i=1}^N \left ( \E[X_i ^2] + 2\sum_{i,j} \E[X_i]\E[X_j]\right ) \right] \\ &&&= \E \left [ N(\sigma^2 + \mu^2) + (N^2-N)\mu^2\right] \\ &&&= n(\sigma^2+\mu^2) + \left ( \frac{n^2-n}{3}-n \right)\mu^2 \\ &&&= n\sigma^2 + \frac{n^2-n}{3} \mu^2 \\ \Rightarrow && \var[Y] &= n\sigma^2 + \frac{n^2-n}{3} \mu^2 - n^2\mu^2 \\ &&&= n\sigma^2 - \frac{2n^2+n}{3} \mu^2 \end{align*}

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Solution: \begin{align*} && \E[H1] &= \sum_{r=1}^n r \cdot \mathbb{P}(\text{first success on }r\text{th turn}) \\ &&&= \sum_{r=1}^n r \cdot q^{r-1}p \\ &&&= p\sum_{r=1}^n r q^{r-1} \\ \\ && \frac{1-x^{n+1}}{1-x} &= \sum_{r=0}^n x^r \\ \Rightarrow && \sum_{r=1}^n r x^{r-1} &= \frac{-(n+1)x^n(1-x) +(1-x^{n+1})}{(1-x)^2} \\ &&&= \frac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2} \\ \\ && \E[H1] &= p\sum_{r=1}^n r q^{r-1} \\ &&&= p\frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} \\ &&&= p^{-1}(1-(n+1)q^{n} + nq^{n+1}) \end{align*} Not that if \(n =0\) , the formula for \(\E[H1] = 0\). So \begin{align*} && \E[H2] &= \E[\E[H1|n=N]] \\ &&&= p^{-1}\E \left [ 1-(N+1)q^{N} + Nq^{N+1}\right] \\ &&&= p^{-1}\E \left [ 1-((1-q)N+1)q^{N} \right] \\ &&&= p^{-1}\left (1 - p\E[Nq^N] - G_{Po(\lambda)}(q) \right) \\ &&&= p^{-1}(1-e^{-\lambda(1-q)}) - \E[Nq^N] \\ &&&= p^{-1}(1-e^{-\lambda(1-q)}) - q\lambda e^{-\lambda(1-q)} \\ &&&= p^{-1}(1-e^{-\lambda p}) - q\lambda e^{-\lambda p} \end{align*}

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Solution:

1. 

   + - ↺
   \begin{align*} \mathbb{E}(A \mid \text{exactly one point on diameter}) &= \int_{-1}^1\int_0^\pi \frac12 (x-0)\cdot 1 \cdot \sin(\pi - \theta) \frac{1}{\pi} \d \theta \frac{1}{2} \d x \\ &= \int_{-1}^1\frac1{2\pi} x \d x \cdot \left [ -\cos \theta \right]_0^\pi \\ &= \frac{1}{2\pi} \end{align*}
2. 

   + - ↺
   \begin{align*} \mathbb{E}(A \mid \text{no point on diameter}) &= \int_0^{\pi} \frac12 \cdot 1 \cdot 1 \cdot \sin x \cdot 2(\pi - x)/\pi^2 \d x \\ &= \frac1{\pi^2} \int_0^\pi \sin x (\pi - x) \d x \\ &= \frac1{\pi^2} \int_0^\pi x\sin x \d x \\ &= \frac1{\pi^2} \left [ \sin x - x \cos x \right]_0^{\pi} \\ &= \frac{1}{\pi} \end{align*}

If both points lie on the diameter the area of the triangle is \(0\). Therefore: \begin{align*} \mathbb{E}(A) &= \frac{1}{2\pi} \mathbb{P}(\text{exactly one point on diameter}) + \frac{1}{\pi}\mathbb{P}(\text{no points on diameter}) \\ &= \frac1{2\pi} \cdot \left (2 \cdot \frac{2}{2+\pi} \cdot \frac{\pi}{2+\pi} \right) + \frac{1}{\pi} \cdot \left ( \frac{\pi}{2+\pi} \cdot \frac{\pi}{2+\pi}\right) \\ &= \frac{1}{\pi} \frac{2\pi + \pi^2}{(2+\pi)^2} \\ &= \frac{1}{2+\pi} \end{align*}