X and Y are two random variables for which joint pdf is given by P(x=i, y=j) = c (i+j) i=1, 2, 3, 4; j=1, 2, 3 find c and conditional mean and variance of X given Y=1.

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written 8.5 years ago by

teamques10 ★ 68k

• modified 8.5 years ago

X and Y are discrete RVs

To find c:

We can tabulate the probabilities as follows:

P( x=i,y=j ) = c( i + j) 1≤x≤4,1≤y≤3

X╲Y	1	2	3	Total
1	2c	3c	4c	9c
2	3c	4c	5c	12c
3	4c	5c	6c	15c
4	5c	6c	7c	18c
Total	14c	18c	22c	54c

Since $∑p_i=1$

$∴ 54c=1$

$∴c=\frac1{54}$

With this value the probability distribution is

X╲Y	1	2	3	Total
1	$\frac2{54}$	$\frac3{54}$	$\frac4{54}$	$\frac9{54}$
2	$\frac3{54}$	$\frac4{54}$	$\frac5{54}$	$\frac{12}{54}$
3	$\frac4{54}$	$\frac5{54}$	$\frac6{54}$	$\frac{15}{54}$
4	$\frac5{54}$	$\frac6{54}$	$\frac7{54}$	$\frac{18}{54}$
Total	$\frac{14}{54}$	$\frac{18}{54}$	$\frac{22}{54}$	1

To find:

E(X/Y=1),Var(X/Y=1)

The marginal probability distribution of Y is:

Conditional Probabilities of X are

$$P(X=1/Y=1)=\frac{P(X=1,Y=1)}{P(Y=1)} =\frac{\frac{\frac{2}{54}}{14}}{54}=\frac{1}{7}$$

$$P(X=2/Y=1)=\frac{P(X=2,Y=1)}{P(Y=1)} =\frac{\frac{(3/54)}{14}}{54}=3/14$$

$$P(X=3/Y=1)=P(X=3,Y=1)/P(Y=1) =\frac{(\frac{(4/54)}{14})}{54}=2/7$$

$$P(X=4/Y=1)=\frac{P(X=4,Y=1)}{P(Y=1)} =\frac{((5/54)/14)}{54}=5/14$$

Now

$E(X/Y=1)=∑x_i p(x=i/y=1)$

=$\frac17+\frac6{14}+\frac67+\frac{20}{14}$

$E(X/Y=1)=\frac{20}7$

$E(\frac{X^2}{Y=1})=∑x_i^2 p(x=i/y=1)=\frac17+\frac{12}{14}+\frac{18}7+\frac{80}{14}$

$E(X^2/Y=1)=\frac{65}7$

$Var(X/Y=1)=E(X^2/Y=1)-{E(X/Y=1)}^2$

$=\frac{65}7-(\frac{20}7)^2$

=9.2857-8.1633

Var($\frac{X}{Y}=1$)=1.1224

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