Published: Thu 21 April 2016
By Saket Choudhary
In RSS-Diploma-Module1 .
Paper
Paper Link
Problem 1
Problem (i)
\begin{align*}
F(x) &= P(X \leq x\\
&= \sum_{A} P(X \leq x, A)\\
&= P(X \leq x, A) + P(X \leq x, A^c)\\
&= P(X \leq x|A)P(A) + P(X \leq x|A^c)P(A^c)\\
&= F_1(x)\theta + F_2(x)(1-\theta)
\end{align*}
To deduce the next equation, we simply
differentiate the above:
\begin{align*}
\frac{d}{dx} F(x) &= \frac{d}{dx} F_1(x)\theta+F_2(x)(1-\theta)\\
f(x) &= f_1(x)\theta+f_2(x)(1-\theta)\\
\int x f(x)dx &= \int x\theta f_1(x)dx + \int x(1-\theta)f_2(x)dx\\
E[X] &= \theta \mu_1 + (1-\theta)\mu_2
\end{align*}
Problem (ii)
\begin{align*}
W_i &\sim \mathcal{N}(4.5,0.25)\\
S_6=\sum_{i=1}^6 W_I &\sim \mathcal{N}(27,1.5)\\
P(S_6\leq 25) &= P(\frac{S_6-27}{\sqrt{1.5}} \leq -\frac{2}{\sqrt{1.5}})\\
&= P(Z \leq -1.63)\\
&=0.056
\end{align*}
Problem (iii)
\begin{align*}
EW &= P(S_6 \leq 25)E[S_7] + P[S_6 \geq 25]E[S_6]\\
&= 0.056*31.5+(1-0.056)*27\\
&= 27.322
\end{align*}
Problem (iv)
\begin{align*}
P(S_6 \leq 25) &= 0.01\\
P(Z \leq -\frac{2}{\sigma}) &= 0.01
\implies -\frac{2}{\sigma}&= -2.33\\
&= 1.165
\end{align*}
Problem 2
Problem (i)
Median = \(\int_{-\infty}^m = \frac{1}{2}\)
Thus,
\begin{align*}
\int_{-\infty}^m xf(x) dx &= \frac{1}{2}\\
\int_{-p}^{\infty}
x f(-x) = \frac{1}{2}\\
&= \int_p^{\infty} xf(x)\\
&= \int_{-p}^{\infty} xf(-x)dx\\
\implies p &=0
\end{align*}
Thus median \(m=0\)
Problem (ii)
\begin{align*}
EX &= \int_{-\infty}^{\infty}xf(x)dx \\
&= \int_{\infty}^{-\infty}-xf(-x)d(-x) \ x \longrightarrow -x\\
&= -\int_{-\infty}^{\infty}xf(-x)dx\\
&= -\int_{-\infty}^{\infty}xf(x)dx \text{ since } f(x)=f(-x) \\
\implies E[X] &=0
\end{align*}
Problem (iii)
\(Y=X^2\)
\begin{align*}
E[XY] &= E[X^3]\\
&= \int_{-\infty}^\infty x^3f(x)dx\\
&=0
\end{align*}
\(Cov(X<y)=E[XY]-E[X]E[Y] = 0\) and hence \(X,Y\) are uncorrelated
Problem (iv)
\begin{align*}
g(y) &= f_X(\sqrt{y}) \frac{dx}{dy}\\
&= \frac{1}{2\sqrt{y}}f_X(\sqrt{y})\\
\text{Note the typo in the original question where $\frac{1}{2}$ is issing}
\end{align*}
Problem 3
Problem 3.(i)
\begin{align*}
f(x,y) &= \frac{1}{x}e^{-x} \\
f(x) &= \int_0^{x} f(x,y)dy\\
&= e^{-x}\\
f_{Y|X}(x)&=\frac{f(X=x,Y)}{f(X)}\\
&= \frac{1}{x}
\end{align*}
Problem 3.(ii)
\begin{align*}
E[X^mY^n] &= \int_{y}^ix^my^n\frac{1}{x}e^{-x} dxdyi\\
&= \int_0^\infty x^{m-1e^{-xdx} \int_0^x y^mdy}\\
&= \int_0^\infty x^{m-1+n+1}e^{-x}\frac{1}{n}\\
&= \frac{(m+n)!}{n+1}
\end{align*}
Problem 3.(iii)
\begin{align*}
E[X]&=1\\
E[X^2] &= 2\\
Var(X) &= 1\\
E[Y]&= 1/2\\
E[Y^2] &= 2/3\\
Var(Y) &= 5/12
\end{align*}
Problem 4
Problem 4.(i)
\begin{tabular}{|c|c|c|}
X1/X2 & 0 & 1\\
0 &01. &0.1\\
1 & 0.3 &0.5
\end{tabular}
Problem 4.(ii)
\begin{align*}
S_Y &= \{0, 1, 2, 3\}\\
\end{align*}
Idea: Start of with a random row and column in the table,
Then keep sampling until there are 10 samples(Ignoring anything that is not in \(S_Y\) )
Problem 4.(iii)
This is straightforward based on theses caseSs
\begin{align*}i
Y&=0 \implies X_1=0,X_2=0\\
Y&=1 \implies X_1=0,X_2=1\\
Y&=2 \implies X_1=1, X_2=0\\
Y&=3 \implies X_1=1,X_2=1
\end{align*}
Problem 4b
TODO
Problem 5
Problem 5.(i)
\begin{align*}
cor(X_1,X_2) &= \frac{Cov(X_1,X_2)}{\sigma_1\sigma_2}\\
&= \frac{2}{4}\\
&= 1/2
\end{align*}
Problem 5.(ii)
\begin{align*}
E[Y] &= -1+1 =0\\
Var(Y) &= Var(X_1) + Var(X_2) + 2Cov(X_1,X_2)\\
&= 21
\end{align*}
Problem 5.(ii)
Since \(Y_1.Y_2\) are both normal, independnce requires just the covariance being zero.
\begin{align*}
Cov(Y_1,Y_2) &=0\\
&= KVar(X_1)+ (k+1)Cov(X_1,X_2) + Var(X_2)\\
&= 3k+18\\
\implies k &=-6
\end{align*}
Problem 5.(b)
\(E[\mathbf{Y}]=\mathbf{0}\)
\(V=\begin{pmatrix}
1/3 & 2/9 & 1/9 & 0 & \dots & 0\\
2/9 & 2/9 & 1/9 & 0 & \dots & 0\\
1/9 & 1/9 & 1/9 & 0 & \dots & 0\\
0 & 0 & 0 & 1/3 & 2/9 & 1/9\\
\end{pmatrix}
\)
Essentiall th 3 rows block shifts every 4th row.
Joint distribution of \(Y_1,Y_2,\dots, Y_{n-2}\) follows a MVN with mean 0 and variance as the above matrix
Problem 6
\begin{align*}
M_X(t) &= E[e^{tX}]\\
&= \int_0^\infty \frac{e^{tx} \theta^\alpha x^{\alpha-1} e^{-\theta x}}{\Gamma(\alpha)}dx\\
&= \int_0^\infty \frac{(\theta-t)^\alpha}{(\theta-t)^\alpha} \frac{ \theta^\alpha x^{\alpha-1} e^{t-\theta x}}{\Gamma(\alpha)}\\
&= \frac{\theta^\alpha}{(t-\theta)^\alpha}\\
EX &= M_X'(0)\\
&= \alpha\theta^\alpha(\theta-t)^{-\alpha-1}\\
&= \alpha/\theta\\
EX^2 &= M_X''(0)\\
&= \alpha(\alpha-1)/\theta^2\\
Var(X) = \alpha/\theta^2-\alpha(\alpha-1)/\theta^2\\
&= \alpha/\theta^2
\end{align*}
Problem 6.(ii)
\begin{align*}
M_X(t) &= \frac{\theta}{\theta-t}
Z &= \sum_iX_i\\
M_Z(t) &= \prod M_{X_i}(t)\\
&= (\frac{\theta}{\theta-t})^n\\
&\sim \Gamma(\theta,n)
\end{align*}
Problem 6.(iii)
Central Limit Theorem:
If \(X_i\) represent random variables whose mgf exits in a neighborhood of 0
and has finite first and second momements, then \(\frac{\bar{X_n}-\mu}{\sigma/\sqrt{n}} \sim \mathcal{N}(0,1)\)
\(\Gamma(n,\theta) \sim \mathcal{N}(n\sqrt{n}/\theta,n^2/\theta^2 )\)
Problem 7
\begin{align*}
P[X_2=1] &= \sum_{_{x_1}} P(X_2=1|X_=x_1)P(X_1=x__1)\\
&= \frac{m}{n}*\frac{m-1}{n-1} + \frac{n-m}{n}\frac{m}{n-1}\\
&= \frac{m}{n}\\
P[X_3=1] &= \sum_{_{x_1,x_2}} P(X-_2=1X_2=x_2,X_1=x_1)\\
\dots
\end{align*}
Problem 8
