joint distribution

• Apr 12th 2007, 11:21 PM
joint distribution
Hi I have two problems on which I am stuck:

In a vehicle servicing system, a vehicle arrives and is immediately placed in a service queue. It may be served immediately or it may have to wait in the queue. After waiting in the queue service, it is serviced and leaves immediately after being serviced. Let X be the tatal time, in minutes, the vehicle is in the system, waiting time plus service time. Let Y be the waiting time in the service queue. Assume X and Y are continuous random variable, where their bivariate space is{0<=y<=x and 0<=x}
Assume the pdf of X and Y is:

-f(x,y)=k*e^(x^2) , 0<=y<=x and 0<=x
-0 elsewhere

1- Find k.
2- Find the conditional pdf of Y given X=x

3- Find the expected value of Y given x=25.

The problem is that f(x,y) does not seem to have y as variable. I found it difficult to work it .

Thank you
B
PROBLEM 2

Find the probability the gas station revenue is at least \$6000( assume normality)

please can you give me some hints?
B
• Apr 13th 2007, 07:01 AM
CaptainBlack
Quote:

Hi I have two problems on which I am stuck:

In a vehicle servicing system, a vehicle arrives and is immediately placed in a service queue. It may be served immediately or it may have to wait in the queue. After waiting in the queue service, it is serviced and leaves immediately after being serviced. Let X be the tatal time, in minutes, the vehicle is in the system, waiting time plus service time. Let Y be the waiting time in the service queue. Assume X and Y are continuous random variable, where their bivariate space is{0<=y<=x and 0<=x}
Assume the pdf of X and Y is:

-f(x,y)=k*e^(x^2) , 0<=y<=x and 0<=x
-0 elsewhere

1- Find k.

The region over which f(x,y) is non-zero is (x>=0) and (0<=y<=x)

then for f to be a pdf we require that:

integral_{x=0 to infty} integral_{y=0 to x} f(x,y) dy dx = 1

or:

integral_{x=0 to infty} integral_{y=0 to x} k e^(x^2) dy dx = 1

but this integral does not converge so presumably what was meant was:

f(x,y) = k e^(-x^2),

then

integral_{x=0 to infty} integral_{y=0 to x} k e^(-x^2) dy dx = k/2,

so as this must equal 1, we have k=2.

RonL
• Apr 13th 2007, 07:04 AM
CaptainBlack
Quote:

Hi I have two problems on which I am stuck:

In a vehicle servicing system, a vehicle arrives and is immediately placed in a service queue. It may be served immediately or it may have to wait in the queue. After waiting in the queue service, it is serviced and leaves immediately after being serviced. Let X be the tatal time, in minutes, the vehicle is in the system, waiting time plus service time. Let Y be the waiting time in the service queue. Assume X and Y are continuous random variable, where their bivariate space is{0<=y<=x and 0<=x}
Assume the pdf of X and Y is:

-f(x,y)=k*e^(x^2) , 0<=y<=x and 0<=x
-0 elsewhere

1- Find k.
2- Find the conditional pdf of Y given X=x

If X=x then Y ~ U(0,x)

Quote:

3- Find the expected value of Y given x=25.
and E(Y|X=25) = 12.5

RonL
• Apr 13th 2007, 07:08 AM
CaptainBlack
Quote:

PROBLEM 2

Find the probability the gas station revenue is at least \$6000( assume normality)

please can you give me some hints?
B

There is not enough information to do this.

RonL
• Apr 14th 2007, 07:18 AM
thank you CaptainBlack.

But what do you mean by:
Quote:

If X=x then Y ~ U(0,x)
?
• Apr 14th 2007, 07:43 AM
CaptainBlack
Quote:

thank you CaptainBlack.

But what do you mean by:
?

If the rv X takes value x then the conditional distribution of Y is the uniform
distribution on (0,x).

RonL
• Apr 15th 2007, 11:56 AM
CaptainBlack,

Please can you be more explicit.
Code:

` E(Y|X=25) = 12.5`
I tried but I cannot find it.

Also for the same problem,
How can I compute the marginal pdf of X

I would do int{x=0 to x=infinty}2e^(-x^2)dx=1.77
is it correct?
B.
• Apr 15th 2007, 12:49 PM
CaptainBlack
Let's hope I've done this right (needless to say this is not how I did it
originaly - it help if you can spot what the answer has to be)

E(Y|X=25) = integral_{y=0 to 25} y P(Y=y|X=25) dy

but P(Y=y|X=25)P(X=25) = P(Y=y, X=25) = k e^{-25^2}

but P(X=25) = integral_{y=0 to 25} k e^{-25^2} dy = k 25 e^{-25^2}

so P(Y=y|X=25) = 1/25 for y in (0,25), and so:

E(Y|X=25) = integral_{y=0,25} y/25 dy = (25^2/2)/25 = 12.5

RonL
• Apr 15th 2007, 12:52 PM
CaptainBlack
Quote: