Hi there. I am aware of how to do the calculation of the following problem, but for the integration, I am not sure how to find the limits.
Suppose (X,Y) is a bivariate continuous random variable with joint probability density function
f(x,y)= 2exp(-x-y) for 0
0 otherwise
Work out P(X+Y > 3 given that X<2).
This therefore means:
P(X<2, X+Y>3) / P(X<2)
For P(X<2) the upper limit would be 2 and the lower limit 0
For P(X<2, X+Y>3)
It is
the double integral of f(x,y)dxdy + the integral of fx(x)dx
I would be grateful for any help on how to work out the limits.
The solution says the limit of the double integral is first (upper limit 3/2, lower limit 0) and then (upper limit infinity, lower limit 3-x)
The second integral has (upper limit 2, lower limit 3/2)
I would be most grateful for any help.
Thank you
Suppose (X,Y) is a bivariate continuous random variable with joint probability density function
f(x,y)= 2exp(-x-y) for 0
Work out P(X+Y > 3 given that X<2).
This therefore means:
P(X<2, X+Y>3) / P(X<2)
For P(X<2) the upper limit would be 2 and the lower limit 0
For P(X<2, X+Y>3)
It is
the double integral of f(x,y)dxdy + the integral of fx(x)dx
I would be grateful for any help on how to work out the limits.
The solution says the limit of the double integral is first (upper limit 3/2, lower limit 0) and then (upper limit infinity, lower limit 3-x)
The second integral has (upper limit 2, lower limit 3/2)
I would be most grateful for any help.
Thank you
-
For P(X < 2, X + Y > 3), the region or integration is above the lines x + y = 3, and x = y for x in [0, 2].
(Sketch them!!) Note that the lower boundaries of the region change where these lines intersect (at x = 3/2). So, we can break this region into 2 parts:
(i) y = 3 - x for x in [0, 3/2]
(ii) y = x for x in [3/2, 2].
So, P(X < 2, X + Y > 3)
= ∫(x = 0 to 3/2) ∫(y = 3 - x to ∞) 2e^(-x-y) dy dx + ∫(x = 3/2 to 2) ∫(y = x to ∞) 2e^(-x-y) dy dx
(Rewrite the integrand as e^(-x) e^(-y) for easier integration...)
---------------------------------
More easily, P(X < 2) = ∫(x = 0 to 2) ∫(y = x to ∞) 2e^(-x-y) dy dx.
I hope this helps!
(Sketch them!!) Note that the lower boundaries of the region change where these lines intersect (at x = 3/2). So, we can break this region into 2 parts:
(i) y = 3 - x for x in [0, 3/2]
(ii) y = x for x in [3/2, 2].
So, P(X < 2, X + Y > 3)
= ∫(x = 0 to 3/2) ∫(y = 3 - x to ∞) 2e^(-x-y) dy dx + ∫(x = 3/2 to 2) ∫(y = x to ∞) 2e^(-x-y) dy dx
(Rewrite the integrand as e^(-x) e^(-y) for easier integration...)
---------------------------------
More easily, P(X < 2) = ∫(x = 0 to 2) ∫(y = x to ∞) 2e^(-x-y) dy dx.
I hope this helps!