I came across the following stats problem online and I was wondering if my approach in solving this problem is correct. Could you kindly offer me any guidance?
Problem statement.
Assume that $X_1$ $\sim$ N(4, 9), $X_2$ $\sim$ B(16, 0.25), $X_3$ $\sim$ $\chi^2$(6) and $X_4$ has a beta distribution with $\alpha$ = 4 and $\beta$ = 6. The random variables $X_1$, $X_2$ and $X_3$ are independent. Furthermore we see that Cov($X_1$,$X_4$) = 0.4, Cov($X_2$,$X_4$) = −0.2 and Cov(X3,X4) = 0.15.
- Give the distribution of the following random variable:
$$W=\left(\frac{X_1-4}{3}\right)^2\;+X_3$$
Current Approach (to verify if correct).
- Find the join probability of $X_1$ and $X_3$
Given that $X_1$ $\perp$ $X_3$, their joint probability function is given by: $f_{X_1,X_3}(X_1,X_3)=f_{X_1}(X_1)f_{X_3}(X_3)$.
- Transform to variable $W$
If we want to substitute to $X_1$, we will transform to $W$ $$ X_1=3\sqrt{W-X_3}+4 $$
$$ f_W(W)=\int f_{X_1}(X_1)f_{X_3}(X_3)\;dX_3 $$
$$ f_W(W)=\int^W_0 f_{X_1}(3\sqrt{W-X_3}+4)f_{X_3}(X_3)\;dX_3 $$
The distributions are given by:
$$ f_{X_3}(X_3)=\frac{1}{\Gamma(\frac{v}{2})2^{v/2}}x_3^{v/2-1}e^{-x_3/2} $$
$$ f_{X_1}(3\sqrt{W-X_3}+4)=\frac{1}{\sigma\sqrt{2\pi}}e^{-(3\sqrt{W-X_3}+4-\mu)^2 /2\sigma^2} $$
The $f_W(W)$ is proportional to the following integral (short way to not write the constants before the integral as it takes a bit of space):
$$ f_W(W)\propto \int^W_0 x_3^{v/2-1}e^{-x_3/2} e^{-(3\sqrt{W-X_3}+4-\mu)^2 /2\sigma^2}dx_3 $$
Then, I think I need to simplify the integral by plugging $\mu$, and $v$,
$$ f_W(W)\propto e^{-(w+25)/18}\int^W_0 e^{-(8x_3-30\sqrt{w-x_3})}x_3^5 dx_3 $$
After here, I am lost: I think I need to integrate this until I reach the form of a standard normal distribution that integrates to 1 or $\Gamma(x)$ function but I am not sure. Any input is welcome!
mmm the easiest way to obtain the requested distribution is by recognizing things and applying a moment generating function. How you are doing it turns out to be very hard and tedious. Note that $X_1$ and $X_3$ are independent. And $$ Z = \frac{X_1-4}{3} \sim N(0,1) $$ and $$ Z^2 = \left( \frac{X_1-4}{3} \right)^2 \sim \chi_{(1)}^{2} $$ and we have $$ W = Z^2 + X_3 $$ the sum of two $\chi_{(\cdot)}^2$ independent. Using moment generating function we have \begin{align*} M_{W}(w) & = M_{Z^2}(w)M_{X_3}(w) \\ & = \frac{1}{(1-2w)^{1/2}} \frac{1}{(1-2w)^{6/2}} \\ & = \frac{1}{(1-2t)^{7/2}} \end{align*} then we have $$ W \sim \chi_{(7)}^{2} $$