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foofighters26
September 11th 2005, 11:03 PM
I am having trouble with the following question, can anyone help me?

Let x1 and x2 be independent random variables each having a uniform dustribution on the interval between 0 and k,so fx(x) = 1/k. Set y=min{x1,x2) and z=max{x1,x2}

Show that the conditional distribution of Y given Z=z is uniform on the interval between 0 and z.

zhaohuihan
September 12th 2005, 02:29 AM
The solution is in the attachment!

Zhaohui

sachin
September 12th 2005, 03:21 PM
P(Y=y/Z=z)=1{x1>x2} P(x2=y/x1=z)+ 1{x1 =< x2} P(x1=y/x2=z)
=1{x1>x2} 1/y +1{x1 =< x2} 1/y ofcourse if 0<y<k
=1/y

Here P(x1=y/x2=z)= P(x1=y) since x1 and x2 are independent.
1{x1<x2) =1 if x1<x2
=0 else

Zhaohui Han
September 12th 2005, 10:58 PM
That is not correct. Your answer is a function of y, isn't it? It's supposed to be a function of z.

sachin
September 13th 2005, 03:30 AM
Ya you are right... i did one little mistake.
P(Y=y/Z=z)=1{x1>x2} P(x2=y/x1=z)+ 1{x1 =< x2} P(x1=y/x2=z)
=1{x1>x2} 1/z +1{x1 =< x2} 1/z ofcourse if 0<y<z
=1/z
Since
P(x1=y/x2=z)=P(x1=y) x1 can be atmost z i.e 0<x1<z
=1/z

1{x1<x2) =1 if x1<x2
=0 else

So Y is Uniform on (0 z)

Zhaohui Han
September 13th 2005, 09:13 AM
I don't think the second equality in your deviations is right, since x1 and x2 are continuous random variables, so that their prob at a single point is zero, instead of 1/z.
How do you think?

sachin
September 13th 2005, 05:09 PM
No that is not a big problem..
Let me explain briefly.

P(Y<= y/Z=z)=P(x1<x2).P(Y<=y/(Z=z, x1<x2)) + P(x1>= x2).P(Y<=y/(Z=z, x1>=x2))

I think that is correct with probability 1..
Now
....P(x1<x2)=1/2
....P(x1>= x2)=1/2
P(Y<=y/(Z=z, x1<x2))=P(Y<=y,Z=z, x1<x2)/P(Z=z, x1<x2)
=P(x1<=y,x2=z, x1<x2)/P(x2=z, x1<x2)
=P(x1<=y, x1<z)/P( x1<z)
=y/k / z/k if y<z
1 if y>=z

i.e
P(Y<=y/(Z=z, x1<x2))=y/z if y<z
= 1 if y>=z
similarly
P(Y<=y/(Z=z, x1>=x2))=y/z if y<z
= 1 if y>=z

So,

P(Y<= y/Z=z)=y/z if y<z
= 1 if y>=z
This is nothing but U(o,z).
Now?

sachin
September 13th 2005, 05:42 PM