Joint, Continuous Random Variables Question

[smartin19]

Homework Statement

Let X and Y have the joint probability density function f(x,y)=k(1-y), if 0<x<y<1 and 0 elsewhere.

a)Find the value of k that makes this pdf valid.
b) Find P(X<3/4,Y>1/2)
c) Find the marginal density function of X and Y
d) Find the expected value and variance of X and Y
e) Find the correlation coefficient
f) Are X and Y independent?

Homework Equations

∫∫f(x,y)dxdy=1
marginal pdf of x=∫f(x,y)dy
marginal pdf of y=∫f(x,y)dx
correlation=covariance(x,y)/(σx * σy)
covariance(x,y)=[∫∫xyf(x,y)dxdy]-[μx*μy]

The Attempt at a Solution

I'm fairly confident I did part A correctly (k=6 is my answer). For part B I set up the double integral (in this case the two double integrals based on the way the region was split) and came up with P(X<3/4, Y>1/2)=∫∫6-6y dydx (1/2<y<1, 0<x<1/2) +∫∫6-6y dydx (x<y<1, 1/2<x<3/4). After evaluating I found P(X<3/4, Y>1/2)≈0.484. Part C seemed fairly straight forward, plug the function into the formulas to get the marginal pdf's; having done that I got f(x)=3-6x+3x^2=pdf of x and f(y)=6y-6y^2=pdf of y. However, when we did examples like this in class, to find the expectation and variance of each variable we found the distribution of each variable based on the marginal pdf's having a form similar to the pdf of some common distribution. I don't see what distributions my marginal pdf's would follow, and, hence, I don't know how to get the expectations.
Thanks

 

[Ray Vickson]

Homework Statement

Let X and Y have the joint probability density function f(x,y)=k(1-y), if 0<x<y<1 and 0 elsewhere.

a)Find the value of k that makes this pdf valid.
b) Find P(X<3/4,Y>1/2)
c) Find the marginal density function of X and Y
d) Find the expected value and variance of X and Y
e) Find the correlation coefficient
f) Are X and Y independent?

Homework Equations

∫∫f(x,y)dxdy=1
marginal pdf of x=∫f(x,y)dy
marginal pdf of y=∫f(x,y)dx
correlation=covariance(x,y)/(σx * σy)
covariance(x,y)=[∫∫xyf(x,y)dxdy]-[μx*μy]

The Attempt at a Solution

I'm fairly confident I did part A correctly (k=6 is my answer). For part B I set up the double integral (in this case the two double integrals based on the way the region was split) and came up with P(X<3/4, Y>1/2)=∫∫6-6y dydx (1/2<y<1, 0<x<1/2) +∫∫6-6y dydx (x<y<1, 1/2<x<3/4). After evaluating I found P(X<3/4, Y>1/2)≈0.484. Part C seemed fairly straight forward, plug the function into the formulas to get the marginal pdf's; having done that I got f(x)=3-6x+3x^2=pdf of x and f(y)=6y-6y^2=pdf of y. However, when we did examples like this in class, to find the expectation and variance of each variable we found the distribution of each variable based on the marginal pdf's having a form similar to the pdf of some common distribution. I don't see what distributions my marginal pdf's would follow, and, hence, I don't know how to get the expectations.
Thanks

NEVER, NEVER use the same letter (f) for two different functions (actually, three different functions in this case). Use different symbols, such as g(x) and h(y) for the marginal densities of X and Y, or use subscripts, like this $f_X(x), f_Y(y)$.

Anyway, you have the marginal pdf of X (but I did not check it!), so finding the expected value and variance of X is just a matter of applying the standard formulas/definitions. I am not grasping what your problem is here. Is it that your $f_X(x)$ does not fit in with some standard, named and tabulated distribution from a list of standard distributions? If so, that does not matter: there are general formulas for the mean and variance for *any* pdf, and all you need do is evaluate the expressions.

 

[smartin19]

Wow, I am completely clueless sometimes. Sorry about the function notation, typically I use subscripts, but apparently I wasn't paying attention. Thanks for your help and quick response.