Expected Value - Probability - COVARIANCE

100 Views Asked by At

I need some help about calculating a covariance from the joint density function.

The density function is:

$$f_{X,Y}(x,y)=\frac{1}{x}\cdot\exp\left(-\frac{4x^2+y}{2x}\right)\cdot I_{x>0, y>0}$$

I need to calculate $E[XY]$ and I don't know how. I know how to calculate $E[X]$ and $E[Y]$ using that function, but not $E[XY]$.

1

There are 1 best solutions below

5
On BEST ANSWER

Calculate

$$E(XY)=\int_{0}^{\infty}\int_{0}^{\infty}xy.f_{X,Y}(x,y)dxdy$$

$$=\int_{x=0}^{\infty}\int_{y=0}^{\infty}y.e^{-\frac{4x^{2}+y}{2x}}dxdy$$

$$=\int_{x=0}^{\infty}e^{-2x}dx\int_{y=0}^{\infty}y.e^{-\frac{y}{2x}}dy$$

$$=\int_{x=0}^{\infty}4x^{2}e^{-2x}dx\int_{z=0}^{\infty}ze^{-z}dz$$putting, $\frac{y}{2x}=z$.

$$=\frac{1}{2}\int_{u=0}^{\infty}u^{2}e^{-u}du=1$$ using the fact $\int_{0}^{\infty}ze^{-z}dz=1 $ & again put $2x=u$.

Further $$COV(X,Y)=E(XY)-E(X)E(Y)$$

I think it helps you. If you want more details tell me.