Suppose f(x,y) is a prob density function: f(x,y) = 1/k -2<=x<y<=2
Find the value of k such that f(x,y) is a valid probability distribution function?
Find the marginal density of X and the marginal density of Y?
Thanks
To be a valid PDF the integral has to equal 1 right? I have an answer but I don't know if it is right!
k= 1+0.5x-0.25x^2. I integrated with respect to first x and then y. The first integral i got x/k and then subbed in y and -2. Then i integrated again and got y^2/2k - 2y/k and then input limits of 2 and x and ended up with the above!
That makes no sense. The integral over an area cannot be a function of either variable. The "outer" integral must have constants as limits. Your region is described as "-2<=x<y<=2" which is the same as saying $\displaystyle -2\le x\le 2$, $\displaystyle x\le y\le 2$, which would give an integral $\displaystyle \int_{x= -2}^2 \int_{y= x}^2 \frac{1}{k} dydx$. Or it can be given as $\displaystyle -2\le y\le 2$, $\displaystyle -2\le y\le x$, which would give an integral $\displaystyle \int_{y= -2}^2\int_{x= -2}^y \frac{1}{k}dxdy$.
A simpler way to do this is to note that since k is a constant, the integral $\displaystyle \int\int \frac{1}{k}dydx= \frac{1}{k}\int\int dA= \frac{1}{k}A$. That is, the integral of just dxdy (or dydx) over a region in the plane is the area of that region. Here the region is a triangle of with vertices at (-2, -2), (-2, 2), and (2, 2). That triangle is a right triangle with legs of length 4 and so has area (1/2)(4)(4)= 8. Your condition is 8/k= 1.
The "marginal density", of X, is the probatility of x over all possible values of y. It is given by the first integral $\displaystyle \int_{y= x}^2 \frac{1}{k}dy$, a function of x. The marginal density of Y is the other way: $\displaystyle \int_{x= -2}^y \frac{1}{k} dx$, a function of y.