# Obtaining f_x for E(X)

• Jul 10th 2011, 10:05 PM
DenJansen
Obtaining f_x for E(X)
I'm need some assistance with intermediate information for a problem I'm attempting to solve. For

$\displaystyle f(x,y)=\left\{ \begin{array}{cc} K(x^{2}+y^{2}) & 20\leq x\leq30,20\leq y\leq30\\0 & \textrm{otherwise}\end{array}\right.$

I'm attempting to solve for E(X) . As per the definition,

$\displaystyle E(X)=\int_{20}^{30}x\cdot f_{x}(x)dx$

Where my confusion lies is, based on my calculations,

$\displaystyle f_{x}=2Kx$

But, according to my solutions manual (corroborated by other sources),

$\displaystyle f_{x}=10Kx^{2}+.05$

I don't understand how they obtained these values. What am I missing here?
• Jul 10th 2011, 10:30 PM
Abu-Khalil
Re: Obtaining f_x for E(X)
By definition, for $\displaystyle 20\le x\le 30$

$\displaystyle f_X(x)=\int_{-\infty}^\infty f_{X,Y}(x,y)dy=\int_{20}^{30}K(x^2+y^2)dy=10Kx^2+ \frac{500}3K$

So

$\displaystyle \mathbb E(X)=\int_{-\infty}^\infty xf_X(x)dx=\int_{20}^{30}x\left(10Kx^2+\frac{500}3K \right)dx$

Maybe you should find the value of $\displaystyle K$.
• Jul 10th 2011, 10:38 PM
DenJansen
Re: Obtaining f_x for E(X)
Wait, I missed the definition of the MPDFs in the last section. So

$\displaystyle \begin{array}{cc}f_{X}(x)=\int_{-\infty}^{\infty}f(x,y)dy & \textrm{for }-\infty<x<\infty\end{array}$

Therefore,

$\displaystyle f_{X}(x)=K\int_{20}^{30}(x^{2}+y^{2})dy=K\left[yx^{2}+\frac{1}{3}y^{3}\right]_{20}^{30}=10Kx^{2}+.05$

where $\displaystyle K=\frac{3}{380,000}$.

NOW it makes sense!