How Do You Calculate Conditional Probabilities with Joint PDFs?

[franz32]

I just need a guide to this problem... found in one of the books in the library...
Given the joint pdf f(x,y) = 2e^[-(x+y)] where 0 < x < y, y > 0

find P(Y < 1 / x < 1). Note that "/" means given that.

I got the formula when P(a < Y < b / X = x) is given, i.e., in terms of the integral from a to b of f(y/x)dy. But how about this? Is there a formula to transform this? =) The answer is in the book... so I want to know how to begin with...

 

[ehild]

I just need a guide to this problem... found in one of the books in the library...
Given the joint pdf f(x,y) = 2e^[-(x+y)] where 0 < x < y, y > 0

find P(Y < 1 / x < 1). Note that "/" means given that.

I got the formula when P(a < Y < b / X = x) is given, i.e., in terms of the integral from a to b of f(y/x)dy. But how about this? Is there a formula to transform this? =) The answer is in the book... so I want to know how to begin with...

I think the probability density function was given as

f(x,y) = 2e^[-(x+y)] where 0 < x < y, y > 0
and f(x,y)=0 if x>y.

Let be the event A: y<1 and the event B: x<1.
By the definition of conditional probability P(A/B) = P(AB)/P(B). (AB means A AND B). You have to integrate f(x,y) to get the probabilities. Find the integration domains in the picture: for P(AB), it is 0<x<y, 0<y<1. For P(B), it is x<y<infinity, 0<x<1.

ehild

 

[M98Ranger]

To solve this problem, we can use the definition of conditional probability, which states that P(A/B) = P(A and B)/P(B). In this case, we are trying to find P(Y < 1 / X < 1), so we can rewrite this as P(Y < 1 and X < 1) / P(X < 1).

To find the joint probability P(Y < 1 and X < 1), we can integrate the joint pdf f(x,y) over the region where both x and y are less than 1. This region can be represented as a triangle with vertices at (0,0), (1,0), and (1,1). So the integral becomes:

∫∫ f(x,y) dx dy = ∫∫ 2e^[-(x+y)] dx dy = 2∫∫ e^[-(x+y)] dx dy

Since the integral is over a triangle, we can change the limits of integration to make the calculations easier. Let u = x+y and v = y, then the triangle becomes a rectangle with vertices at (0,0), (1,0), (1,1). The limits of integration become 0 to 1 for both u and v. The Jacobian of this transformation is 1, so the integral becomes:

2∫∫ e^[-u] du dv = 2∫ e^[-u] dv = 2e^-u ∫ dv = 2e^-u (v) = 2e^-u (1-0) = 2e^-u

Now, to find P(X < 1), we can integrate the joint pdf f(x,y) over the region where x is less than 1. This region can be represented as a triangle with vertices at (0,0), (1,0), and (1,∞). So the integral becomes:

∫∫ f(x,y) dx dy = ∫∫ 2e^[-(x+y)] dx dy = 2∫∫ e^[-(x+y)] dx dy

Again, we can change the limits of integration to make the calculations easier. Let u = x+y and v = y, then the triangle becomes a trapezoid with vertices at (0,0), (1,0), (1,∞), and (0,