Problems

Solution:

1. Note that \(\displaystyle F(y) = \mathbb{P}(X_i < y) = \int_0^y \lambda e^{-\lambda t} \d t = 1-e^{-\lambda y}\). Notice also that \begin{align*} G(y) &= \mathbb{P}(Y < y) \\ &= \mathbb{P}(\max_i(X_i) < y) \\ &= \mathbb{P}(X_i < y \text{ for all }i) \\ &= \prod_{i=1}^n \mathbb{P}(X_i < y) \\ &= \prod_{i=1}^n (1-e^{-\lambda y})\\ &= (1-e^{-\lambda y})^n \end{align*} as required.
2. \begin{align*} && \mathbb{P}(Y < L(\alpha)) &= \alpha \\ \Rightarrow && (1-e^{-\lambda L(\alpha)})^n &= \alpha \\ \Rightarrow && 1-e^{-\lambda L(\alpha)} &= \alpha^{\tfrac1n} \\ \Rightarrow && L(\alpha) &= -\frac{1}{\lambda}\ln \left (1-\alpha^{\tfrac1n} \right) \end{align*} Notice also: \begin{align*} && \mathbb{P}(Y > U(\alpha)) &= \alpha \\ \Rightarrow && 1 - (1-e^{-\lambda U(\alpha)})^n &= \alpha \\ \Rightarrow && U(\alpha) &= -\frac{1}{\lambda}\ln \left ( 1-(1-\alpha)^{\tfrac1n} \right) \end{align*}
3. \begin{align*} \lambda L(\alpha) &= -\ln \left (1-\alpha^{\tfrac1n} \right) \\ &= -\ln \left (1-e^{\tfrac1n \ln \alpha} \right) \\ &\approx - \ln \left ( 1 - 1 - \frac1n \ln \alpha\right) \tag{\(e^t \approx 1 + t\)} \\ &= -\ln \left ( \frac{1}{n} \ln \frac{1}\alpha \right) \\ &= - \ln \frac{1}{n} - \ln \left ( \ln \frac{1}{\alpha} \right )\\ &= \ln n - \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right) \end{align*} since if \(n\) is large, \(\frac{\ln \alpha}{n}\) is small.
4. The median is the value where \(\mathbb{P}(Y < M) = \frac12\), or in other words \(L(\frac12)\), but this is \(\approx \frac{\ln n - \ln (\ln 2)}{\lambda} \to \infty\). \begin{align*} && \lambda U(\alpha) &\approx \ln n - \ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right) \\ \Rightarrow && \lambda(U(\alpha) - L(\alpha)) &\approx -\ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right)+ \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right) \\ \Rightarrow && U(\alpha) - L(\alpha) &\to \frac{1}{\lambda} \left ( \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right)-\ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right ) \right) \end{align*} which doesn't depend on \(n\).
5. Suppose \(\alpha = \frac{1}{20}\) then \begin{align*} U(\alpha) - L(\alpha) &\approx \frac{1}{\lambda} \left (\ln \ln 20 - \ln \ln \frac{20}{19} \right) \\ &= \lambda^{-1} \left (\ln \ln 20 - \ln \ln (1 + \frac{1}{19}) \right) \\ &\approx \lambda^{-1} \left (\ln 3 - \ln \frac{1}{19} \right) \tag{\(\ln(1+t) \approx t\)} \\ &\approx \lambda^{-1} \ln 3 \cdot 19 \\ &\approx \lambda^{-1} (1 + 3) \\ &\approx 4\lambda^{-1} \end{align*} [Note that \(\ln \ln 20 - \ln \ln \frac{20}{19} = 4.0673\ldots\)]

Solution:

1. \(\,\) \begin{align*} && 1 &= \int_0^1 kx^n(1-x) \d x \\ &&&= k\left [\frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_0^1 \\ &&&= \frac{k}{(n+1)(n+2)} \\ \Rightarrow && k &= (n+1)(n+2) \\ \\ && \E[X] &= \int_0^1 kx^{n+1}(1-x) \d x \\ &&&= k\left [ \frac{x^{n+2}}{n+2} - \frac{x^{n+3}}{n+3} \right]_0^1 \\ &&&= \frac{(n+1)(n+2)}{(n+2)(n+3)} \\ &&&= \frac{n+1}{n+3} \end{align*}
2. If \(\mu\) is less than the median then \begin{align*} && \frac12 &> \int_0^{\mu} kx^{n}(1-x) \d x \\ &&&= \left [k \frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_0^\mu \\ &&&= \mu^{n+1}\left ((n+2) - (n+1) \frac{n+1}{n+3} \right) \\ \Rightarrow && \left ( 1 + \frac{2}{n+1} \right)^{n+1} &> 2 \frac{(n+2)(n+3) - (n+1)^2}{n+3} \\ &&&= \frac{6n+10}{n+3} = 6 - \frac{8}{n+3} \end{align*} \begin{align*} && \left ( 1 + \frac{2}{n+1} \right)^{n+1} &= 1 + (n+1) \frac{2}{n+1} + \frac{(n+1)n}{2} \frac{4}{(n+1)^2} + \frac{(n+1)n(n-1)}{6} \frac{8}{(n+1)^3} + \cdots \\ &&&= 1 + 2 + \frac{2n}{n+1} + \frac{4n(n-1)}{3(n+1)^2} + \cdots \\ &&&= 5 - \frac{2}{n+1} + \frac{4((n+1)^2-3(n+1)+3)}{3(n+1)^2} \\ &&&= 6 + \frac13 - \frac{6}{n+1} - \frac{4}{3(n+1)^2} \\ &&&> 6 - \frac{8}{n+3} \end{align*} as required.
3. To find the mode, we need to find the maximum of \(x^n(1-x)\) which occurs when \(nx^{n-1}(1-x) - x^n = x^{n-1}(n-(n+1)x)\) ie where \(x = \frac{n}{n+1}\), therefore we need to look at: \begin{align*} && \mathbb{P}(X \leq \tfrac{n}{n+1}) &= \int_0^{n/(n+1)} k x^n(1-x) \d x \\ &&&= \frac{n^{n+1}}{(n+1)^{n+1}} \left ( (n+2) - (n+1) \frac{n}{n+1} \right) \\ &&&= 2\frac{n^{n+1}}{(n+1)^{n+1}} \\ &&&= 2 \left ( \frac{1}{1+\frac1n} \right)^{n+1} \\ &&&\geq 2 \frac{1}{(1 + \frac11)^2} = \frac12 \end{align*} as required

Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(Y = n) &= \mathbb{P}(X \in [n, n+1)) \\ &&&= \int_n^{n+1} \lambda e^{-\lambda x} \d x \\ &&&= \left [-e^{-\lambda x} \right]_n^{n+1} \\ &&&= e^{-\lambda n} - e^{-\lambda(n+1)} \\ &&&= e^{-\lambda n}(1- e^{-\lambda}) \end{align*}
2. \(,\) \begin{align*} && \mathbb{P}(Z < z) &= \sum_{i=0}^{\infty} \mathbb{P}(X \in (n, n+z)) \\ &&&= \sum_{i=0}^{\infty} \int_{n}^{n+z} \lambda e^{-\lambda x} \d x \\ &&&= \sum_{i=0}^{\infty} [-e^{-\lambda x}]_{n}^{n+z} \\ &&&= \sum_{i=0}^{\infty} (1-e^{-\lambda x})e^{-\lambda n} \\ &&&= \frac{1-e^{-\lambda x}}{1-e^{-\lambda}} \end{align*}
3. Give the cdf of \(Z\), we see that \(f_Z(z) = \frac{\lambda e^{-\lambda z}}{1-e^{-\lambda}}\) so \begin{align*} && \E[Z] &= \int_0^1 z \frac{\lambda e^{-\lambda z}}{1-e^{-\lambda}} \d z \\ &&&= \frac{\lambda}{1-e^{-\lambda}} \int_0^1 ze^{-\lambda z} \d z \\ &&&= \frac{\lambda}{1-e^{-\lambda}} \left ( \left [-\frac{1}{\lambda} ze^{-\lambda z} \right]_0^1+\int_0^1 \frac{1}{\lambda} e^{-\lambda z} \d z \right) \\ &&&= \frac{\lambda}{1-e^{-\lambda}} \left ( -\frac{e^{-\lambda}}{\lambda} + \frac{1-e^{-\lambda}}{\lambda^2} \right) \\ &&&= \frac{1-e^{-\lambda}(1+\lambda)}{\lambda (1-e^{-\lambda})} \end{align*}
4. \(\,\) \begin{align*} && \mathbb{P}(Y = n \text{ and }z_1 < Z < z_2)&= \mathbb{P}(X \in (n+z_1, n+z_2) ) \\ &&&= \int_{n+z_1}^{n+z_2} \lambda e^{-\lambda x} \d x \\ &&&= e^{-n\lambda}(e^{-\lambda z_1} - e^{-\lambda z_2}) \end{align*} Note that \(\mathbb{P}(z_1 < Z < z_2) = \mathbb{P}( Z < z_2) -\mathbb{P}(Z< z_1) =\frac{e^{-\lambda z_1} - e^{-\lambda z_2}}{1-e^{-\lambda}}\) Therefore \begin{align*} && \mathbb{P}(Y = n \text{ and }z_1 < Z < z_2) &= e^{-n\lambda}(e^{-\lambda z_1} - e^{-\lambda z_2}) \\ &&&= e^{-\lambda n}(1-e^{-\lambda}) \frac{e^{-\lambda z_1} - e^{-\lambda z_2}}{1-e^{-\lambda}} \\ &&&= \mathbb{P}(Y=n) \mathbb{P}(z_1 < Z < z_2) \end{align*} So they are independent, which is to be expected from the memorylessness property of the exponential distribution.

Solution:

1. \(\,\) \begin{align*} && \mu &= \E[X] \\ &&&= \int_0^1 x f(x) \d x \\ &&&= \int_0^1 nx^n \d x \\ &&&= \frac{n}{n+1} \\ \\ && \var[X] &= \sigma^2 \\ &&&= \E[X^2] - \mu^2 \\ &&&= \int_0^1 x^2 f(x) \d x - \mu^2 \\ &&&= \int_0^1 nx^{n+1} \d x - \mu^2 \\ &&&= \frac{n}{n+2} - \frac{n^2}{(n+1)^2} \\ &&&= \frac{n(n+1)^2 - n^2(n+2)}{(n+1)^2(n+2)} \\ &&&= \frac{n}{(n+1)^2(n+2)} \end{align*}
2. \(\,\) \begin{align*} && \frac14 &= \int_0^{Q_1} 2x \d x \\ &&&= Q_1^2 \\ \Rightarrow && Q_1 &= \frac12 \\ && \frac34 &= \int_0^{Q_3} 2x \d x \\ &&&= Q_3^2 \\ \Rightarrow && Q_3 &= \frac{\sqrt{3}}2 \\ \\ \Rightarrow && IQR &= Q_3 - Q_1 = \frac{\sqrt{3}-1}{2} \\ && 2 \sigma &= 2\sqrt{\frac{2}{3^2 \cdot 4}} \\ &&&= \frac{\sqrt{2}}{3} \\ \\ && 2\sigma - IRQ &= \frac{\sqrt{2}}{3} - \frac{\sqrt{3}-1}{2} \\ &&&= \frac{2\sqrt{2}-3\sqrt{3}+3}{6} \\ && (3+2\sqrt{2})^2 &= 17+12\sqrt{2} > 29 \\ && (3\sqrt{3})^2 &= 27 \end{align*} Therefore \(2\sigma > IQR\)
3. \[ (1+x)^n = 1 + nx + \frac{n(n-1)}2 x^2 + \cdots + \binom{n}{k} x^k+ \cdots \] \begin{align*} && Q_1^{-n} &= 4 \\ && Q_2^{-n} &= 2\\ && \mu &=\frac{n}{n+1} \\ \Rightarrow && \mu^{-n} &= \left (1 + \frac1n \right)^n\\ &&&\geq 1 + n \frac1n + \cdots > 2 \\ \Rightarrow && \mu &< Q_2 \\ \\ && \mu^{-n} &= \left (1 + \frac1n \right)^n\\ &&&= 1 + n \frac1n + \frac{n(n-1)}{2!} \frac{1}{n^2} + \cdots + \frac{n(n-1) \cdots (n-k+1)}{k!} \frac{1}{n^k} + \cdots \\ &&&= 1 + 1 + \left (1 - \frac1n \right ) \frac1{2!} + \cdots + \left (1 - \frac1n \right)\cdot\left (1 - \frac2n \right) \cdots \left (1 - \frac{k-1}n \right) \frac{1}{k!} + \cdots \\ &&&< 1 + 1 + \frac1{2!} + \cdots + \frac1{k!} \\ &&&< 4 \\ \Rightarrow && \mu &> Q_1 \end{align*}

Solution:

1. \begin{align*} && \mathbb{P}(Y_k \leq y) &= \sum_{j=k}^n\mathbb{P}(\text{exactly }j \text{ values less than }y) \\ &&&= \sum_{j=k}^m \binom{m}{j} y^j(1-y)^{n-j} \end{align*}
2. This is the number of ways to choose a committee of \(m\) people with the chair from those \(m\) people. This can be done in two ways. First: choose the committee in \(\binom{n}{m}\) ways and choose the chair in \(m\) ways so \(m \binom{n}{m}\). Alternatively, choose the chain in \(n\) ways and choose the remaining \(m-1\) committee members in \(\binom{n-1}{m-1}\) ways. Therefore \(m \binom{n}{m} = n \binom{n-1}{m-1}\) \begin{align*} (n-m) \binom{n}{m} &= (n-m) \binom{n}{n-m} \\ &= n \binom{n-1}{n-m-1} \\ &= n \binom{n-1}{m} \end{align*} \begin{align*} f_{Y_k}(y) &= \frac{\d }{\d y} \l \sum^{n}_{m=k}\binom{n}{m}y^{m}\left(1-y\right)^{n-m} \r \\ &= \sum^{n}_{m=k} \l \binom{n}{m}my^{m-1}\left(1-y\right)^{n-m} -\binom{n}{m}(n-m)y^{m}\left(1-y\right)^{n-m-1} \r \\ &= \sum^{n}_{m=k} \l n \binom{n-1}{m-1}y^{m-1}\left(1-y\right)^{n-m} -n \binom{n-1}{m} y^{m}\left(1-y\right)^{n-m-1} \r \\ &= n\sum^{n}_{m=k} \binom{n-1}{m-1}y^{m-1}\left(1-y\right)^{n-m} -n\sum^{n+1}_{m=k+1} \binom{n-1}{m-1} y^{m-1}\left(1-y\right)^{n-m} \\ &= n \binom{n-1}{k-1} y^{k-1}(1-y)^{n-k} \end{align*} \begin{align*} &&1 &= \int_0^1 f_{Y_k}(y) \d y \\ &&&= \int_0^1 n \binom{n-1}{k-1} y^{k-1}(1-y)^{n-k} \d y \\ &&&= n \binom{n-1}{k-1} \int_0^1 y^{k-1}(1-y)^{n-k} \d y \\ \Rightarrow && \frac{1}{n \binom{n-1}{k-1}} &= \int_0^1 y^{k-1}(1-y)^{n-k} \d y \\ \end{align*}
3. \begin{align*} && \mathbb{E}(Y_k) &= \int_0^1 y f_{Y_k}(y) \d y \\ &&&= \int_0^1 n \binom{n-1}{k-1} y^{k}(1-y)^{n-k} \\ &&&= n \binom{n-1}{k-1}\int_0^1 y^{k}(1-y)^{n-k} \d y \\ &&&= n \binom{n-1}{k-1}\int_0^1 y^{k+1-1}(1-y)^{n+1-(k+1)} \d y \\ &&&= n \binom{n-1}{k-1} \frac{1}{(n+1) \binom{n}{k}}\\ &&&= \frac{n}{n+1} \cdot \frac{k}{n} \\ &&&= \frac{k}{n+1} \end{align*}

Solution: \begin{align*} {\rm V}(x) &= \E \big( (X-x)^2\big) \\ &= \E \l X^2 - 2xX + x^2\r \\ &= \E [ X^2 ]- 2x\E[X] + x^2 \\ &= \sigma^2+\mu^2 - 2x\mu + x^2 \\ &= \sigma^2 + (\mu - x)^2 \end{align*} \begin{align*} \E[Y] &= \E[\sigma^2 + (\mu - X)^2] \\ &= \sigma^2 + \E[(\mu - X)^2]\\ &= \sigma^2 + \sigma^2 \\ &= 2\sigma^2 \end{align*} If \(X \sim U(0,1)\) then \(V(x) = \frac{1}{12} + (\frac12 - x)^2\). \begin{align*} \P(Y \leq y) &= \P(\frac1{12} + (\frac12 - X)^2 \leq y) \\ &= \P((\frac12 -X)^2 \leq y - \frac1{12}) \\ &= \P(|\frac12 -X| \leq \sqrt{y - \frac1{12}}) \\ &= \begin{cases} 1 & \text{if } y - \frac1{12} > \frac14 \\ 2 \sqrt{y - \frac1{12}} & \text{if } \frac14 > y - \frac1{12} > 0 \\ \end{cases} \\ &= \begin{cases} 1 & \text{if } y> \frac13 \\ \sqrt{4y - \frac1{3}} & \text{if } \frac13 > y > \frac1{12} \\ \end{cases} \end{align*} Therefore $f_Y(y) = \begin{cases} \frac{2}{\sqrt{4y-\frac{1}{3}}} & \text{if } \frac1{12} < y < \frac13 \\ 0 & \text{otherwise} \end{cases}$ \begin{align*} \E[Y] &= \int_{1/12}^{1/3} \frac{2x}{\sqrt{4x-\frac13}} \, dx \\ &= 2\int_{u = 0}^{u=1} \frac{\frac{1}{4}u +\frac1{12}}{\sqrt{u}} \,\frac{1}{4} du \tag{\(u = 4x - \frac13, \frac{du}{dx} = 4\)}\\ &= \frac{1}{2 \cdot 12}\int_{u = 0}^{u=1} 3\sqrt{u} +\frac{1}{\sqrt{u}} \, du \\ &= \frac{1}{2 \cdot 12} \left [2 u^{3/2} + 2u^{1/2} \right ]_0^1 \\ &= \frac{1}{2 \cdot 12} \cdot 4 \\ &= \frac{2}{12} \end{align*} as required

Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(T > t) &= \mathbb{P}(\text{all emails slower than }t) \\ &&&= \left ( \int_t^{\infty} \lambda e^{-\lambda x} \d x \right)^n \\ &&&= \left ( [- e^{-\lambda x}]_t^\infty\right)^n\\ &&&= e^{-n\lambda t} \\ \Rightarrow && f_T(t) &= n \lambda e^{-n\lambda t} \\ \end{align*} Therefore \(T \sim \text{Exp}(n \lambda)\) and \(\E[T] = \frac{1}{n \lambda}\)
2. Let \(T_2\) be the time until the second email arrives, then. \begin{align*} && \P(T_2 > t) &= \P(\text{all emails} > t) + \P(\text{all but 1 emails} > t) \\ &&&= e^{-n\lambda t} + n \cdot e^{-(n-1)\lambda t}(1-e^{-\lambda t}) \\ &&&= (1-n)e^{-n\lambda t} + n \cdot e^{-(n-1)\lambda t} \\ \Rightarrow && f_{T_2}(t) &= - \left ( (1-n) n \lambda e^{-n \lambda t} -n(n-1)\lambda e^{-(n-1)\lambda t} \right) \\ &&&= n(n-1) \lambda \left (e^{-(n-1)\lambda t} - e^{-n\lambda t} \right) \\ \Rightarrow && \E[T_2] &= \int_0^{\infty} t \cdot n(n-1) \lambda \left (e^{-(n-1)\lambda t} - e^{-n\lambda t} \right) \d t \\ &&&= \int_0^{\infty} \left (n \cdot t (n-1) \lambda e^{-(n-1)\lambda t} -(n-1)\cdot tn \lambda e^{-n\lambda t} \right) \d t \\ &&&= \frac{n}{\lambda(n-1)} - \frac{n-1}{\lambda n} \\ &&&= \frac{1}{\lambda} \left (1+\frac{1}{n-1}- \left (1 - \frac{1}{n} \right) \right) \\ &&&= \frac{1}{\lambda} \left ( \frac{1}{n-1} + \frac{1}{n} \right) \end{align*} (We can also view this second expectation as expected time for first email + expected time (of the remaining \(n-1\) emails) for the first email, and we can see that will have that form by the memorilessness property of exponentials)

Solution:

1. \(\mathbb{P}(X + Y \leq c) \) is the area between the \(x\)-axis, \(y\)-axis and the line \(x + y = c\). There are two cases for this: \[\mathbb{P}(X + Y \leq c) = \begin{cases} 0 & \text{ if } c \leq 0 \\ \frac{c^2}{2} & \text{ if } c \leq 1 \\ 1- \frac{(2-c)^2}{2} & \text{ if } 1 \leq c \leq 2 \\ 1 & \text{ otherwise} \end{cases}\]
2. \begin{align*} && \mathbb{P}((X + Y)^{-1} \leq t) &= 1- \mathbb{P}(X + Y \leq \frac1{t}) \\ \Rightarrow && f_{(X+Y)^{-1}}(t) &= 0 -\begin{cases} 0 & \text{ if } \frac1{t} \leq 0 \\ \frac{\d}{\d t}\frac{1}{2t^2} & \text{ if } \frac{1}{t} \leq 1 \\ \frac{\d}{\d t} \l 1- \frac{(2-\frac1t)^2}{2} \r & \text{ if } 1 \leq \frac{1}{t} \leq 2 \\ 0 & \text{ otherwise}\end{cases} \\ && &= \begin{cases} t^{-3} & \text{ if } t \geq 1 \\ (2-\frac1t)t^{-2} & \text{ if } \frac12 \leq t \leq 1\\ 0 & \text{ otherwise}\end{cases} \\ && &= \begin{cases} t^{-3} & \text{ if } t \geq 1 \\ 2t^{-2}-t^{-3} & \text{ if } \frac12 \leq t \leq 1\\ 0 & \text{ otherwise}\end{cases} \end{align*} Therefore, \begin{align*} \E \Big(\dfrac1{X+Y}\Big) &= \int_{\frac12}^{\infty} t f_{(X+Y)^{-1}}(t) \, \d t \\ &= \int_{\frac12}^{1} t f_{(X+Y)^{-1}}(t) \, \d t + \int_{1}^{\infty} t f_{(X+Y)^{-1}}(t) \d t\\ &= \int_{\frac12}^{1} \l 2t^{-1} - t^{-2} \r \, \d t + \int_{1}^{\infty} t^{-2} \d t\\ &= \left [ 2 \ln (t) + t^{-1} \right]_{\frac12}^{1} + \left [ -t^{-1} \right ]_{1}^{\infty} \\ &= 1 + 2 \ln 2 -2 + 1 \\ &= 2 \ln 2 \end{align*}
3. \begin{align*} &&\mathbb{P} \l \frac{Y}{X} \leq c \r &= \mathbb{P}( Y \leq c X) \\ &&&= \begin{cases} 0 & \text{if } c \leq 0 \\ \frac{c}{2} & \text{if } 0 \leq c \leq 1 \\ 1-\frac{1}{2c} & \text{if } 1 \leq c \end{cases} \\ \\ \Rightarrow && \mathbb{P} \l \frac{X}{X+Y} \leq t\r &= \mathbb{P} \l \frac{1}{1+\frac{Y}{X}} \leq t\r \\ &&&= \mathbb{P} \l \frac{1}{t} \leq 1+\frac{Y}{X}\r \\ &&&= \mathbb{P} \l \frac{1}{t} - 1\leq \frac{Y}{X}\r \\ &&&= 1- \mathbb{P} \l \frac{Y}{X} \leq \frac{1}{t} - 1\r \\ &&&= 1 - \begin{cases} 0 & \text{if } \frac1{t} \leq 0 \\ \frac{1}{2t} - \frac{1}{2} & \text{if } 0 \leq \frac1{t} \leq 1 \\ 1-\frac{t}{2-2t} & \text{if } 1 \leq \frac1{t} \end{cases} \\ && f_{\frac{X}{X+Y}}(t) &= \begin{cases} 0 & \text{if } \frac1{t} \leq 0 \\ \frac{1}{2t^2} & \text{if } t \geq 1 \\ \frac{1}{2(1-t)^2} & \text{if } 0 \leq t \leq 1 \end{cases} \\ \Rightarrow && \mathbb{E} \l \frac{X}{X+Y} \r &= \int_0^\infty t f(t) \d t \\ &&&= \int_0^1 \frac{1}{2(1-t)^2} \d t + \int_1^\infty \frac{1}{t^2} \d t \\ &&& = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \\ \\ && \mathbb{E} \l \frac{X}{X+Y} \r &= \int_0^1 \int_0^1 \frac{x}{x+y} \d y\d x \\ &&&= \int_0^1 \l x \ln (x+1) - x \ln x \r \d x \\ &&&= \left [\frac{x^2}2 \ln(x+1) - \frac{x^2}{2} \ln(x) \right]_0^1 -\int_0^1 \l \frac{x^2}{2(x+1)} - \frac{x}{2} \r \d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \int_0^1 \frac{x^2-1+1}{2(x+1)}\d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \int_0^1 \frac{x -1}{2} + \frac{1}{2(x+1)}\d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \frac{1}{4} + \frac{1}{2} - \frac{\ln 2}{2} \\ &&&= \frac{1}{2} \end{align*} We can also notice that \(1 = \mathbb{E} \l \frac{X+Y}{X+Y} \r = \mathbb{E} \l \frac{X}{X+Y} \r + \mathbb{E} \l \frac{Y}{X+Y} \r = 2 \mathbb{E} \l \frac{X}{X+Y} \r\) so it's clearly true as long as we can show that the integral converges.

Solution: Since \(\int f(x) \, dx = 1\), and \(f(x)\) is a triangle with base \(b-a\), it must have height \(\frac{2}{b-a}\) in order to have the desired area. Since \(g(a) = 0, g(c) = \frac{2}{b-a}\), \(g(x) = A(x-a)\) and \(\frac{2}{b-a} = A (c-a) \Rightarrow g(x) = \frac{2(x-a)}{(b-a)(c-a)}\) as required. Similarly, \(h(x) = B(x-b)\) and \(\frac{2}{b-a} = B(c-b) \Rightarrow h(x) = \frac{2(b-x)}{(b-a)(b-c)}\) The mean of the distribution will be: \begin{align*} \int_a^b xf(x) \, dx &= \int_a^c xg(x) \, dx + \int_c^b xh(x) \, dx \\ &= \frac{2}{(b-a)(c-a)} \int_a^c x(x-a) dx + \frac{2}{(b-a)(b-c)} \int_c^b x(b-x) \, dx \\ &= \frac{2}{(b-a)} \l \frac{1}{c-a} \left [ \frac{x^3}{3} - a\frac{x^2}{2} \right ]_a^c + \frac{1}{b-c} \left [ b\frac{x^2}{2} - \frac{x^3}{3} \right ]_c^b\r \\ &= \frac{2}{(b-a)} \l \frac{1}{c-a} \l \frac{c^3}{3} - a\frac{c^2}{2} - \frac{a^3}{3} + \frac{a^3}{2} \r + \frac{1}{b-c} \l \frac{b^3}{2} - \frac{b^3}{3} - \frac{bc^2}{2} + \frac{c^3}{3} \r \r \\ &= \frac{2}{(b-a)} \l \l \frac{c^2+ac+a^2}{3} - \frac{a(a+c)}{2} \r +\l \frac{b(b+c)}{2} - \frac{b^2+bc+c^2}{3} \r\r \\ &= \frac{2}{(b-a)} \l \frac{2c^2+2ac+2a^2}{6} - \frac{3a^2+3ac}{6} + \frac{3b^2+3bc}{6} - \frac{2b^2+2bc+2c^2}{6} \r \\ &= \frac{2}{(b-a)} \l \frac{-a^2+b^2-ac+bc}{6} \r \\ &= \frac{a+b+c}{3} \\ \end{align*} The median \(M\) satisfies: \begin{align*} && \int_a^M f(x) \, dx &= \frac12 \\ \end{align*} The left hand triangle will have area: \(\frac{c-a}{b-a}\) which will be \(\geq \frac12\) if \(c \geq \frac{a+b}{2}\). In this case we need \begin{align*} && \frac{(M-a)^2}{(b-a)(c-a)} &= \frac12 \\ \Rightarrow && M &= a + \sqrt{\frac12 (b-a)(c-a)} \end{align*} Otherwise, we need: \begin{align*} && \frac{(b-M)^2}{(b-a)(b-c)} &= \frac12 \\ \Rightarrow && M &= b - \sqrt{\frac12 (b-a)(b-c)} \end{align*} These are consistent, if \(c = \frac{b+a}{2}\)

Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(T > t + \delta t | T > t) &= \frac{\mathbb{P}(T < t + \delta t)}{\mathbb{P}(T > t )} \\ &&&= \frac{\int_t^{t+\delta t} f(s) \d s}{1-F(t)} \\ &&&\approx \frac{f(t)\delta t}{1-F(t)} \\ &&&= h(t) \delta t \end{align*}
2. If \(F(t) = t/a\) then \(f(t) = 1/a\) and \(h(t) = \frac{1/a}{1-t/a} = \frac{1}{a-t}\)
   

   + - ↺
3. \(\,\) \begin{align*} && \frac{F'}{1-F} &= \frac{1}{t} \\ \Rightarrow && -\ln (1-F) &= \ln t + C\\ \Rightarrow && 1-F &= \frac{A}{t} \\ && F &= 1 - \frac{A}{t} \\ F(a) = 0: && F &= 1 - \frac{a}{t} \\ && f(t) &= \frac{a}{t^2} \end{align*}
4. (\(\Rightarrow\)) \begin{align*} && \frac{F'}{1-F} &= k \\ \Rightarrow && -\ln(1-F) &= kt+C \\ \Rightarrow && 1-F &= Ae^{-kt} \\ F(b) = 0: && 1 &= Ae^{-kb} \\ \Rightarrow && 1-F &= e^{-k(t-b)}\\ \Rightarrow && f &= ke^{-k(t-b)} \\ \end{align*} (\(\Leftarrow\)) \(f(t) = ke^{-k(t-b)} \Rightarrow F(t) = 1-e^{-k(t-b)}\) and the result is clear.
5. \(\,\) \begin{align*} && \frac{F'}{1-F} &= \left ( \frac{\lambda}{\theta^{\lambda}} \right) t^{\lambda-1} \\ \Rightarrow && -\ln(1-F) &= \left ( \frac{t}{\theta} \right)^{\lambda} +C\\ \Rightarrow && F &= 1-A\exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \\ F(0) = 0: && 0 &= 1-A \\ \Rightarrow && F &= 1 - \exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \\ \Rightarrow && f &= \lambda t^{\lambda -1} \theta^{-\lambda} \exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \end{align*}

Solution:

1. \begin{align*} && \frac12 &= \mathbb{P}(X \leq x_m) \\ \Leftrightarrow && \frac12 &= \mathbb{P}(e^X \leq e^{x_m} = y_m) \end{align*} Therefore the median is \(y_m = e^{x_m}\)
2. \begin{align*} && \mathbb{P}(Y \leq y) &= \mathbb{P}(e^X \leq y) \\ &&&= \mathbb{P}(X \leq \ln y) \\ &&&= F(\ln y) \\ \Rightarrow && f_Y(y) &= f(\ln y)/y \\ \\ && f'_Y(y) &= \frac{f'(\ln y) - f(\ln y)}{y^2} \end{align*} Therefore since the mode satisfies \(f'_Y = 0\) we must have \(f'(\ln \lambda ) = f(\ln \lambda)\)
3. This is the integral of the pdf of \(N(\mu + \sigma^2, \sigma^2)\) and therefore is clearly \(1\). \begin{align*} && \E[Y] &= \int_{-\infty}^{\infty} e^x \cdot \frac{1}{\sqrt{2\pi \sigma^2}} e^{-(x-\mu)^2/(2\sigma^2)} \d x \\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (x - (x-\mu)^2/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp ((2x \sigma^2- (x-\mu)^2)/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu-\sigma^2)^2+2\mu \sigma^2-\sigma^4)/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu+\sigma^2)^2)/(2\sigma^2)+\mu +\frac12\sigma^2) \d x\\ &&&= \e^{\mu +\frac12\sigma^2}\frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu-\sigma^2)^2)/(2\sigma^2)) \d x\\ &&&= \e^{\mu +\frac12\sigma^2} \end{align*}
4. Notice that \(y_m = e^\mu < e^{\mu + \tfrac12 \sigma^2} = \E[Y]\), so it suffices to prove that \(\lambda < e^{\mu}\) Notice that \(f'(x) - f(x) = f(x)[-(x-\mu)/\sigma^2 - 1]\) and therefore \(\ln y - \mu = -\sigma^2\) so \(\lambda = e^{\mu - \sigma^2}\) which is clearly less than \(e^{\mu}\) as required.

Solution:

1. 1. \(\,\) \begin{align*} && 0 &\leq f(t) &\leq M \\ \Rightarrow && \int_0^x 0 \d t &\leq \int_0^x f(t) \d t & \leq \int_0^x M \d x \\ \Rightarrow && 0 &\leq F(x) &\leq Mx \end{align*}
   2. \(\,\) \begin{align*} && \int_0^1 2g(x)F(x)f(x) \d x &= \left [ g(x) F(x)^2 \right] - \int_0^1 g'(x) \left ( F(x)\right)^2 \d x \\ &&&= g(1) - \int_0^1 g'(x) \left ( F(x)\right)^2 \d x \end{align*}
2. 1. \(\,\) \begin{align*} && 1 &= \int_0^1 kF(y)f(y) \d y \\ &&&= k\left [ \frac12 F(y)^2\right]_0^1 \\ &&&= \frac{k}{2} \\ \Rightarrow && k &= 2 \end{align*}
   2. \(\,\) \begin{align*} \E[Y^n] &= \int_0^1 y^n 2F(y)f(y) \d y \\ &\geq \int_0^1 y^n 2My f(y) \d y \\ &= 2M\int_0^1 y^{n+1} f(y) \d y \\ &= 2M \E[X^{n+1}] = 2M\mu_{n+1} \\ \\ \E[Y^n] &= \int_0^1 y^n 2F(y)f(y) \d y \\ &= 1 - \int_0^1 ny^{n-1} F(y)^2 \d y \\ &\geq 1 - \int_0^1 ny^{n-1}My F(y) \d y \\ &= 1 - M\int_0^1 ny^n F(y) \d y \\ &= 1 - M[\frac{n}{n+1}y^{n+1} F(y)]_0^1 + M\int_0^1\frac{n}{n+1} y^{n+1} f(y) \d y \\ &= 1 - \frac{nM}{n+1} + \frac{nM}{n+1} \mu_{n+1} \end{align*}
   3. Since \(\E[Y^{n-1}] \geq 0\) we must have \begin{align*} && 2M\mu_n \geq 1 + \frac{(n-1)M}{n}\mu_n - \frac{(n-1)M}{n} \\ \Rightarrow && \mu_n \left (2M + \frac{(n-1)M}{n} \right) \geq 1 - \frac{(n-1)M}{n} \\ \Rightarrow && \mu_n \frac{3Mn-M}{n} & \geq \frac{n-(n-1)M}{n} \\ \Rightarrow && \mu_n & \geq \frac{n-(n-1)M}{3Mn-M} \end{align*}