Joint probability density for independent continuous random variables

In summary: P[B(Y+Z)≤x] = (1-P[B(Y+Z)<=x])^2P[B(Y+Z)≤x] = (1-P[B(Y+Z)<=x])^2 = (1-P[B(Y+Z)<=x])^2 = (1-P[B(Y+Z)<x])In summary, the homework statement states that two continuous random variables, Y and Z, have a common probability density function. The joint probability density function of Y and Z is h(y; z), where y+z is greater than x. E(
  • #1
Kate2010
146
0

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[tex]\leq[/tex] 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 [tex]\geq[/tex] 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 [Broken]
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Kate2010 said:

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 [tex]\leq[/tex] 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 [tex]\geq[/tex] 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 [Broken]
You've done everything right so far. Now you want to calculate

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

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.
 
Last edited by a moderator:
  • #3
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?

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

How's that?
 
  • #4
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.
 
  • #5
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.
 
  • #6
Kate2010 said:
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.
 
  • #7
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.
 
  • #8
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].
 
  • #9
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.
 

1. What is joint probability density?

Joint probability density is a statistical concept that represents the likelihood of two or more random variables occurring simultaneously. It is a function that describes the probability of a set of outcomes for multiple variables, typically denoted as f(x,y) or f(x1,x2,...,xn).

2. How is joint probability density different from marginal probability density?

Joint probability density considers the probability of multiple variables occurring together, while marginal probability density only considers the probability of one variable occurring. In other words, joint probability density takes into account the relationship between variables, while marginal probability density does not.

3. What is the relationship between joint probability density and conditional probability?

Conditional probability is the probability of one event occurring given that another event has already occurred. Joint probability density can be used to calculate conditional probability by dividing the joint probability of two events by the marginal probability of the event that has already occurred.

4. How is joint probability density used in data analysis?

Joint probability density is used in data analysis to understand the relationship between multiple variables and their likelihood of occurring together. It can help identify patterns and correlations between variables, which can be useful in making predictions and drawing conclusions from data.

5. Can joint probability density be greater than 1?

No, joint probability density cannot be greater than 1. This is because it is a measure of probability, which cannot exceed 1. If the joint probability density of two variables is greater than 1, it is likely due to an error in calculation or an incorrect assumption about the variables.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
343
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
695
  • Calculus and Beyond Homework Help
Replies
19
Views
759
  • Calculus and Beyond Homework Help
Replies
3
Views
943
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
Back
Top