Joint probability density for independent continuous random variables

[Kate2010]

Homework Statement

X,Y,Z random variables.

Let Y and Z be two independent continuous random variables with
common probability density function
f(x) =2exp(-2x); x > 0
0; x < 0

(i) Specify the joint probability density function of (Y;Z).

(ii) Fix x > 0. Let
h(y; z) =
1; y + z 6\leq x
0; y + z > x
Calculate E(h(Y,Z)).

Homework Equations

The Attempt at a Solution

i) I don't really understand part i. I think x is a kind of dummy variable but I'm not sure. If it is:

f(y,z)= (2exp(-2y))(2exp(-2z)) = 4exp(-2{y+z}) for y and z \geq 0, and 0 otherwise.

Then E(h(Y,Z)) = the double integral from 0 to infinity both times of h(y,z) * 4exp(-2{y+z}) dy dz

But I'm not totally sure about these limits, or how to sub in h?

Sorry this is very badly typed in and very muddled. Here is the question paper (q8) if this is more helpful. http://www.maths.ox.ac.uk/system/files/private/active/0/paperC2009.pdf

 

[vela]

Homework Statement

X,Y,Z random variables.

Let Y and Z be two independent continuous random variables with
common probability density function
f(x) =2exp(-2x); x > 0
0; x < 0

(i) Specify the joint probability density function of (Y;Z).

(ii) Fix x > 0. Let
h(y; z) =
1; y + z \leq x
0; y + z > x
Calculate E(h(Y,Z)).

Homework Equations

The Attempt at a Solution

i) I don't really understand part i. I think x is a kind of dummy variable but I'm not sure. If it is:

f(y,z)= (2exp(-2y))(2exp(-2z)) = 4exp(-2{y+z}) for y and z \geq 0, and 0 otherwise.

Then E(h(Y,Z)) = the double integral from 0 to infinity both times of h(y,z) * 4exp(-2{y+z}) dy dz

But I'm not totally sure about these limits, or how to sub in h?

Sorry this is very badly typed in and very muddled. Here is the question paper (q8) if this is more helpful. http://www.maths.ox.ac.uk/system/files/private/active/0/paperC2009.pdf

You've done everything right so far. Now you want to calculate

E[h(Y,Z)] = \int_0^\infty \int_0^\infty h(y,z)f(y,z)\,dy\,dz

Geometrically, the region of integration is the first quadrant of the yz plane. But h(y,z) is non-zero over a subset of the quadrant, and that's the region you actually need to integrate over.

 

[Kate2010]

When you say I need to integrate h(y,z), is this multipliedby f(y,z), otherwise I will just be integrating 0 and 1?

\int^{infinity}_{z=-infinity}\int^{x-z}_{y=-infinity} 4exp (-2{y+z}) dy dz

How's that?

 

[vela]

Yes, you're absolutely right. I forgot f(y,z). I fixed the integrand in my previous post.

The limits on your integral aren't quite correct. The function h(y,z) splits the xy plane into two, and the boundary is the line y+z=x. The function f(y,z) limits you to the first quadrant of the yz plane because it's zero when y<0 or z<0. Try sketching this out. It should be straightforward to see what your limits of integration should be then.

 

[Kate2010]

Oh yes I forgot about the y>0 and z>0. However, I tried just changing those to 0 and end up with infinity*exp(-2x) as part of my answer when I integrate. So I don't think I just need to change that.

Is it z = 0 to x, y = 0 to x-z ____ dy dx ?

Also if you have time, could you look at the final part of the question that I linked to? I have no idea how to approach this.

Thank you very much.

 

[vela]

Oh yes I forgot about the y>0 and z>0. However, I tried just changing those to 0 and end up with infinity*exp(-2x) as part of my answer when I integrate. So I don't think I just need to change that.

Is it z = 0 to x, y = 0 to x-z ____ dy dx ?

Yes, though with dz not dx, but that's probably just a typo. Did you sketch the region over which you're integrating? I'm just asking because it sounds a bit like you're just guessing what the limits are rather than figuring out what they should be.

 

[Kate2010]

Oh yes I meant dz rather than dx. I did sketch it but I'm just not very confident with reading from the diagram what the limits should be still. At first I was trying to do a 3d sketch too rather than just taking x as a fixed point of intersection.

 

[vela]

Well, you did get the limits right. Did you get a result for the expected value of h?

For the final part, you want to calculate P[B(Y+Z)≤x]. Consider the cases where B=0 and B=1 separately. In other words, calculate P[B(Y+Z)≤x | B=0] and P[B(Y+Z)≤x | B=1] and combine them appropriately to get P[B(Y+Z)≤x].

 

[Kate2010]

I got (1-2x)exp(-2x) - 1. I may have made calculation errors as I did it in a hurry as my work was due in, but understand the method now, thanks :D

Also, got very close to what I wanted for the final part.

Really appreciate your help.