Let the joint density function of X and Ybe given by:
f(x)={kxy^2 for 0<x<y<1
(0 otherwise
what is the value of k?
Can someone please help me with this?
The formula I used was:
))kxy^2 dx dy (where )) is double integration both from 0 to 1).
I worked it out (I think) and got 15. I used integration by parts where i had u=x, du=dx, dv=y^2 dy, and v=(y^3)/3. If anyone can tell me if this is right, or I'm on the right track, let me know... PLEASE!!!
I'm not sure your integration is correct. This looks like an iterated integral from 0 to y, then from 0 to 1, where you have ))f(x,y)dxdy in that order.
There might be other ways to set this up to, but I'm a little short on time to look at it right now.
Here's the work I did... Sorry if its hard to follow
))kxy^2 dx dy
k))xy^2 dx dy <--Since k is a constant, I can pull it outside the integration sign, right?
k)uv-)v du 0|1 <--that's evaluated from 0 to 1
k)[(xy^3)/3]-[(xy^4)/12] 0|1 <--Only one integration sign, right after the k before the bracket
k)y^3/3-y^4/12
k/3)y^3-y^4/4 dy
k/3(y^4/4-y^5/20) 0|1
k/12- k/60=1
k/15=1
k=15
I really don't understand your integration by parts. What you have is
f(x,y)=kxy2
Let's use the uppercase S for the integral, so
SSD f(x,y)dA would be the integral, where D is the area in the x-y plane we're integrating over.
According to this:
"f(x)={kxy^2 for 0<x<y<1
(0 otherwise
what is the value of k? "
we should integrate from x=0 to x=y.
I also think your f should be a function of both x and y.
Yes, you can pull the k out, so the integral should (if I've read the problem correctly) look like
kSS xy2dxdy where the limits look like
S 10S y0
When you do the first integral you should have
(1/2)x2y2 evalulated from 0 to y for x. This gives
(1/2)y4 if you now evaluate this integral you should get
(1/10)y5 evaluated from 0 to 1 and end up with
k*(1/10)=1 so k=10
Hopefully I've read the problem correctly. Be sure to check your text for a worked example, this is a very common problem and you should have no problem finding an example to compare the problem with.
To find the value of k in the given joint density function, you need to set up and solve the double integral of f(x) over the given limits of integration.
The correct setup of the double integral can be written as:
∫∫ kxy^2 dx dy
To evaluate this integral, you first need to integrate with respect to x from 0 to y, and then integrate with respect to y from 0 to 1.
So, the integral becomes:
∫[0, 1] ∫[0, y] kxy^2 dx dy
To solve this integral, you can integrate the inner integral first (with respect to x) and then the outer integral (with respect to y).
Integrating the inner integral with respect to x gives:
∫[0, y] (kx/3)y^3 dy
Next, you can integrate this with respect to y, using the limits y=0 to y=1:
∫[0, 1] (ky/3)y^3 dy
Integrating gives:
(k/3) ∫[0, 1] y^4 dy
Evaluating this integral from 0 to 1 gives:
(k/3) [(1/5)y^5] evaluated from 0 to 1
So, the integral simplifies to:
(k/3) (1/5)
Setting this equal to 1 (as given in the problem), you get:
(k/3) (1/5) = 1
To solve for k, you can multiply both sides of the equation by 15 (the reciprocal of (1/3) (1/5)):
k = 15
Therefore, the value of k in the joint density function is 15.