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Rup

November 12th 2005, 07:15 AM

Can anybody try to solve and explain these problems? Thanks!

The joint density of the random variables X and Y is ¼ in the square 0<x<2, 0<y<2. Let W=min(X, Y). Calculate E (W).

Answer: 2/3

A random variable X has the following CDF.

F(x)= x/2 0<x<1

1 x>(or equal to) 1

Let M(t) be the moment generating function of X. Calculate M(ln2).

Answer: 1/ln2+1

sszpakowski

November 12th 2005, 07:58 AM

The first problem is a double integral problem. Because the joint density is uniform, you only have to do one integral and multiply the result times 2.

So please forgive my notation, but here's how you're supposed to set up the integral:

2 double integral (x ranging from 0 to 2 and y ranging from 0 to x) of x/4 dydx.

The reason it's only x in the numerator and not xy is because of W=min(X,Y), so in the triangle where we integrated X is the min so W=X.

I hope this helps!

krzysio

November 12th 2005, 03:29 PM

Can anybody try to solve and explain these problems? Thanks!

The joint density of the random variables X and Y is ¼ in the square 0<x<2, 0<y<2. Let W=min(X, Y). Calculate E (W).

Answer: 2/3

Sorry, yall, but you are really trying to do this the most complicated way, unnecessarily so.

E (W) = integral from 0 to 2 of the survival function of W =

= integral from 0 to 2 of (1 - 0.5w)^2 =

= (-2/3)(1 - 0.5w)^3 evaluated from 0 to 2 = 2/3

Simple and direct. I will post this and the other way with my weekly exercises. Neat problem.

Yours,

Krzys'

Rup

November 14th 2005, 04:13 PM

Thanks for your help! :)

Rup

November 15th 2005, 03:33 AM

Mr. Ostaszewski,

I am just wondering did you use ¼ at all in the solution.:confused:

Thanks

Rup

krzysio

November 15th 2005, 06:29 PM

Mr. Ostaszewski,

I am just wondering did you use ¼ at all in the solution.:confused:

Rup

My solution was probably confusing with the typo I had in it, so I corrected it now. So sorry.

No 1/4 in it. I know you wanted 1/4 because it is joint density, but there is no need to us it.

Yours,

Krzys'

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