Question

15. Let X and Y denote the lengths of life, in hundreds of hours, for co ponents of typesI and types II, respectively in an electronic system. The joint density of X and Y is given by Bre" (z +v)/2 f(z, y) = otherwise Find the probability that a component of type II will have a life lenght in excess of 200 hours. 16. Let the random variables X and Y have the joint p.d.f a. f(z,y)=ī, for (z,y) = (0,0), (1,1), (2,2), zero elsewhere b/('·»-|. /or (z, y)-(, 2), (1,1), (2,0), aero elsewhere c.f(r,y)for ()-(0,0), (1, 1), (2,0), ero elsewhere In each case compute the correlation coefficient of X and Y

Answer 1

16. Sol:

(a)

Following table shows the calculations for E(XY):

X	Y	f(x,y)	xyf(x,y)
0	0	0.3333	0
0	1	0.3333	0
2	2	0.3333	1.3332
Total			1.3332

So

$E(XY) = \sum xyf(x,y)=1.3332$

Following table shows the marginal pdf :

X	P(X=x)	xP(X=x)	x^2P(X=x)
0	0.6666	0	0
2	0.3333	0.6666	1.3332
Total		0.6666	1.3332

So,

$E(X)=\sum xP(X=x)=0.6666$

and

$Var(X)=\sum x^{2}P(X=x)-\left [\sum xP(X=x) \right ]^{2}\approx 0.8888$

Following table shows the calculations:

Y	P(Y=y)	yP(Y=y)	y^2P(Y=y)
0	0.3333	0	0
1	0.3333	0.3333	0.3333
2	0.3333	0.6666	1.3332
Total		0.9999	1.6665

So,

$E(Y)=\sum yP(Y=y)=0.9999$

and

$Var(Y)=\sum y^{2}P(Y=y)-\left [\sum yP(Y=y) \right ]^{2}\approx 0.6667$

The covariance is

$Cov(X,Y)=E(XY)-E(X)E(Y)=1.3332- (0.6666\cdot 0.9999)=0.6667$

The correlation coefficient:

$\rho=\frac{Cov(X,Y)}{\sqrt{var(X)Var(Y)}}=0.5132$

(b)

Following table shows the calculations for E(XY):

X	Y	f(x,y)	xyf(x,y)
0	2	0.3333	0
1	1	0.3333	0.3333
2	0	0.3333	0
Total			0.3333

So

$E(XY) = \sum xyf(x,y)=0.3333$

Following table shows the marginal pdf :

X	P(X=x)	xP(X=x)	x^2P(X=x)
0	0.3333	0	0
1	0.3333	0.3333	0.3333
2	0.3333	0.6666	1.3332
Total		0.9999	1.6665

So,

$E(X)=\sum xP(X=x)=0.9999$

and

$Var(X)=\sum x^{2}P(X=x)-\left [\sum xP(X=x) \right ]^{2}\approx 0.6667$

Following table shows the calculations:

Y	P(Y=y)	yP(Y=y)	y^2P(Y=y)
2	0.3333	0.6666	1.3332
1	0.3333	0.3333	0.3333
0	0.3333	0	0
Total		0.9999	1.6665

So,

$E(Y)=\sum yP(Y=y)=0.9999$

and

$Var(Y)=\sum y^{2}P(Y=y)-\left [\sum yP(Y=y) \right ]^{2}\approx 0.6667$

The covariance is

$Cov(X,Y)=E(XY)-E(X)E(Y)=0.3333- (0.9999\cdot 0.9999)=-0.6665$

The correlation coefficient:

$\rho=\frac{Cov(X,Y)}{\sqrt{var(X)Var(Y)}}=-0.4444$

(c)

Following table shows the calculations for E(XY):

X	Y	f(x,y)	xyf(x,y)
0	0	0.3333	0
1	1	0.3333	0.3333
2	0	0.3333	0
Total			0.3333

So

$E(XY) = \sum xyf(x,y)=0.3333$

Following table shows the marginal pdf :

X	P(X=x)	xP(X=x)	x^2P(X=x)
0	0.3333	0	0
1	0.3333	0.3333	0.3333
2	0.3333	0.6666	1.3332
Total		0.9999	1.6665

So,

$E(X)=\sum xP(X=x)=0.9999$

and

$Var(X)=\sum x^{2}P(X=x)-\left [\sum xP(X=x) \right ]^{2}\approx 0.6667$

Following table shows the calculations:

Y	P(Y=y)	yP(Y=y)	y^2P(Y=y)
0	0.6666	0	0
1	0.3333	0.3333	0.3333
Total		0.3333	0.3333

So,

$E(Y)=\sum yP(Y=y)=0.3333$

and

$Var(Y)=\sum y^{2}P(Y=y)-\left [\sum yP(Y=y) \right ]^{2}\approx 0.2222$

The covariance is

$Cov(X,Y)=E(XY)-E(X)E(Y)=0.00003333$

The correlation coefficient:

$\rho=\frac{Cov(X,Y)}{\sqrt{var(X)Var(Y)}}=0.000012828$

As per HomeworkLib policy we need to slove one question per post, please post the remaining question in another post.