improper-integralsintegrationmultivariable-calculusprobabilitystatistics
Joint density function is as follows:
$$ f(x,y) = cx e^{-x(x+y)},\quad \text{ for } x>0, y>0$$ and is equal $0$ elsewhere.
My question is how do I find the normalizing constant c? I have tried working it out by expanding the exponential term and completing the square in hopes of simplifying it by using integrals involving normal pdfs but to no avail… Any solutions?
After Paul's suggestion,
$$c\int_0^{\infty} e^{-x^2} \;\mathrm{dx}$$
How do I deal with $e^{-x^2}$?
$\hskip 1.7 in$
The shadowed region corresponds to all points $(x,y)$ s.t. $f_{X+Y}(x,y)>0$. So, when $0\le a\le 1$ e.g. $a_1$ the region of integration is bounded by the orange line "$a_1-a_1$" i.e. $$\{y>0,x>0,y\le a_1-x\}$$
On the other hand, when $a\ge 1$ e.g. $a_2$ the region of integration is the intersection of the region bounded by the line "$a_2-a_2$" and the shadowed one i.e.
$$\{x>0,y>0,y\le a_2-x,y<1-x/2\}$$
We know that the integral over the density function must be equal to $1$.
That is:
$$\int_0^1 kx^2~ dx+\int_1^2 k(2-x)~ dx=1$$
Computing the two integrals gives:$$\frac{k}{2}+\frac{k}{3}=1 \iff \frac{5k}{6}=1$$ and so we have $$k=\frac{6}{5}$$
Best Answer
The integral over the pdf must be 1, i.e $$ c \cdot \int_0^\infty \int_0^\infty x e^{-x(x+y)} dx dy = 1.$$
Integration over $y$ yields $\frac{1}{x}$ so that it remains to solve the integral $$ \int_0^\infty e^{-x^2} dx.$$ Using the substitution $z=\sqrt{2} \cdot x$, we have to find $$ \frac{1}{\sqrt{2}} \int_0^\infty e^{\frac{-z^2}{2}} dz.$$ The latter integral is - aside from a contant factor - the positive part of the standard normal. All in all we get $$c \cdot \frac{\sqrt{\pi}}{2} = 1 $$ which immediately gives us c.