This is an attempt to go through previous year questions in the paper 1 of Indian statistical service examination
Joint pdf of X and Y be \[f(x,y)=\begin{cases} \begin{align} 6(1-x-y),~~~~~~~ x>0,y>0,x+y<1\\ 0,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Otherwise \end{align} \end{cases} \]
1. What is the value of E(XY)
Sol Since the upper bound is given as \(x+y<1\)
and since lower limit is 0 for both variables
upper bound is 1 for both variables with a little condition x<1-y and y<1-x, we can use 1 as upper bound for any of the variable and any one of previous two values as upper bound for anothe variable.
\[ E(XY)=\int_{0}^{1}\int_0^{1-x}xy6(1-x-y)~dydx \]
Ans D) 1/20
2. What is the \(Cov(X,~Y)\)
Sol
\[Cov(X,~Y)=E(XY)-E(X)E(y)\]
\[f(X)=\int_0^{1-x}6(1-x-y)~ =3x^2-6x+3\] Similarly we can obtain for Y (Which is same with y variable in it)
\[E(X)=\int_{0}^{1}3x^2-6x+3~dx=1/4\] Similarly we can get for Y
\[Cov(X,~Y)=1/20-1/4*1/4\] Ans D) -1/80
3. What is the equation of line of regression of X and Y
Sol
\[(X-\overline{X})=r \frac{\sigma_x}{\sigma_y}(Y-\overline{Y})\]
\[\sigma_x=\sqrt{E(X^2)-[E(X)]^2} \]
\[E(X^2)=\int_0^1x^2(3x^2-6x+3)dx = \frac{1}{10}\]
\[\sigma_x=\sqrt{\frac{1}{10}-\left(\frac{1}{4}\right)^2} =\sqrt{\frac{3}{80}}\]
The values are same for y also
\[\sigma_y=\sqrt{E(Y^2)-[E(Y)]^2} \]
\[\sigma_y=\sqrt{\frac{1}{10}-\left(\frac{1}{4}\right)^2} =\sqrt{\frac{3}{80}}\]
\[r=\frac{COV(XY)}{\sigma_x\sigma_y}=\frac{\frac{-1}{80}}{\sqrt{\frac{3}{80}}\sqrt{\frac{3}{80}}}=\frac{-1~~}{{~}3}\]
The equation \[\left(X-\frac{1}{4}\right)=\frac{-1~~}{{~}3}\left(Y-\frac{1}{4}\right)\]
Ans B) \[X=\frac{1-Y}{3}\]
Consider the following for the next two items that follow
\[ f(x,y)= \begin{cases} 8xy,~~~~ 0<x<y<1\\ 0,~~~~~~~ Otherwise. \end{cases} \]
4. What is \(E[Y|X=x]\) equal to?
Sol \[E[Y|X=x]=\int_{-\infty}^{\infty} y~f(y|x)~dy\]
\[f(x)=\int_x^1 8xy~dy=8x\frac{y^2}{2} \Bigg| _x^1=4x(1-x^2)\]
\[f(y|x)=\frac{f(xy)}{f(y)}=\frac{8xy}{4x(1-x^2)}=\frac{2y}{(1-x^2)}\]
\[E[Y|X=x]=\int_{x}^{1} y~\frac{2y}{(1-x^2)}~dy=\frac{2}{1-x^2}\left(\frac{y^3}{3}\right)\Bigg|_{x}^{1}=\frac{2}{3}\left(\frac{1-x^3}{1-x^2}\right)\]
Ans C
\[\frac{2}{3}\left(\frac{1-x^3}{1-x^2}\right)\]
5. Dropped
6. How large a sample must be taken in order that the probability will be atleast 0.95 and the sample mean will be within 0.5- neighbourhood of the population mean provided population S.D is 1
Sol
\[
\sigma=1\\
\text{Margin of error}=ME\leq0.5\\
P(Z)=0.95\implies Z=1.96\]
\[ME=Z*\frac{\sigma}{\sqrt{n}}\\ ME\leq0.5\\ \frac{1.96*1}{\sqrt{n}} \leq 0.5 \implies n \geq 16\]
Here maximum value in the option is 80 we choose that option
Ans A \[80\]
7. Let \(A_1,A_2,A_3,\dots\) be a sequence of events and let \(E=\lim sup A_n\). If \(\sum_{n=1}^{\infty}P(A_n)<\infty,\) then \(P(E)\) is equal to
Sol
Ans D \[0\]
8.
Consider the following statements in respect of Characteristic function of a random variable:
1. It always exist
2. It is uniformly continuous on R
3.It is not independent of change of origin and scale
4. If characteristic function of sum of two random variables is same as the product of their individual characteristic functions then the variables are independent.
Which of the above staments are correct ?
Sol
Ans A
\[1, 2 \text{ and }3 \text{ only } \]