Derive the distribution of Y using MGF technique

Question

Given the PDF $$f_X(x)=2\lambda^2xe^{-(\lambda x)^2}I_{(0,\infty)}(x); \lambda>0$$
Derive the distribution of $Y=(\lambda x)^2$ using the MGF technique. Here is my attempt:
$$E(e^{t(\lambda x)^2})=\int_{-\infty}^{\infty} e^{t(\lambda x)^2} 2\lambda^2xe^{-(\lambda x)^2} dx$$
$$=\int_{-\infty}^{\infty} e^{-(\lambda x)^2(-t+1)} 2x\lambda^2 dx$$
$$=e^{(-t+1)}\int_{-\infty}^{\infty} e^{-(\lambda x)^2} 2x\lambda^2 dx$$
$$=\lambda e^{(-t+1)}\int_{-\infty}^{\infty} e^{-(\lambda x)^2} 2x\lambda dx$$
It seems to me like the integrand will end up being an exponential distribution with $\lambda=2x$, is that correct? I also don't know what to make of $\lambda e^{(-t+1)}$, it doesn't look like the MGF of any of the distributions to me. Please let me know how to fix my solution.

Answer 1

The integral should be from $0\to\infty$. Because the pdf is non-zero only for positive $x$. So you should do that or while writing the integral from $-\infty\to\infty$ you should also include the $I_{(0,\infty)}$ inside the integral.

$$\int_{0}^{\infty} e^{-(\lambda x)^2(-t+1)} 2x\lambda^2 dx$$

Now you substitute $\lambda^{2}x^{2}(1-t)=z$ so

$2\lambda^{2}xdx=\frac{1}{1-t}dz$

So you get:-

$$\int_{0}^{\infty}\frac{1}{1-t}e^{-z}\,dz=\frac{1}{1-t}$$

Which is the MGF of an $\text{Exp}(1)$ distribution.

So $Y\sim \text{Exp}(1)$

PS:- As an advice, when you are asked to do something by a more complicated method, it is better to first do the problems by easier methods with which you are comfortable with and then proceed with the problem. Then you will know that if the answers don't match up, you might have made a mistake .

For example, it is way easier to just use the Jacobian transformation and say that $Y\sim\text{Exp}(1)$.

To do that you have $Y=X^{2}\lambda^{2}$.

So $\frac{dy}{dx}=2x\lambda^{2}$.

So $|\frac{dx}{dy}|=\frac{1}{2x\lambda^{2}}$

So $f_{Y}(y)=|\frac{dx}{dy}|f_{X}(x)=\frac{1}{2x\lambda^{2}}e^{-(\lambda x)^2} 2x\lambda^{2} I_{(0,\infty)}$

So $f_{Y}(y)=e^{-\lambda^{2}x^{2}}I_{(0,\infty)}=e^{-y}I_{(0,\infty)}$

So you see that the pdf of $Y$ has the same pdf of a $\text{Exp}(1)$ distribution. That lets you easily conclude as it directly gives you the pdf rather than checking which MGF it matches with.