i have a bivariate distribution for (X, Y) with joint density: $$f(x,y) = \frac{\lambda y^2}{\sqrt2\pi}e^-(\frac{1}{2} + \lambda x)y^2$$ on x>0 and y belong to R I need to show that $f_X(x) = \frac{\lambda}{2\sqrt2(\frac{1}{2} + \lambda x)^\frac{3}{2}}$ on x>0
My approach has been this so far but i am stuck! $$f_X(x) =\frac{\lambda y^2}{\sqrt 2\pi}\int_0^x e^\frac{-y}{2}\lambda x$$ $$f_X(x) =\frac{\lambda y^2}{\sqrt 2\pi} \frac{\frac{1}{1}}{\frac{-y}{2}}e^\frac{-y}{2}\lambda x$$ $$f_X(x) =\frac{\lambda y}{\sqrt 2\pi}e^\frac{-y}{2}\lambda x\Big|_0^x$$
I dont know what else to do so i need help please!
$f_X(x)=\int_{\mathbb R} f(x,y)dy$. [Note that $f_X(x)$ can only depend on $x$, not on $y$ so what yo have done is not correct]. To evaluate this integral you have to know how to find $\int_{\mathbb R} e^{-ay^{2}/2}y^{2}dy$ where $a=1+2\lambda x$. This is easy. This integral is just $\frac 1 {a^{3/2}}\sqrt {2\pi}$ from the fact that the second moment of $N(0,\sigma^{2})$ distribution is $\sigma^{2}$. Multiplying this integral by $\frac {\lambda} {\sqrt {2\pi}}$ we get exactly the answer you are suppose to get.