Question

1. Let X₁, X_{2 ,} and X₃ be independent standard normal random variable. If we defined Y₁=X₂, Y₂=X₁+X₂/2 and Y₃=X₁+ X₂ +X₃/3.Then find then joint pdf of Y₁, Y_2, and Y₃ using the Jacobian method?

Answer 1

Suppose that random variables $X_1$, $X_2$, $X_3$ have normal distributions with parameters $\mu_1,\sigma^2_1$, $\mu_2,\sigma^2_2$ and $\mu_3,\sigma^2_3$ respectively. We use a known fact that in case a random variable $Z$ has a normal distribution with parameters $\mu_Z,\sigma^2_Z$, then, a random variable $aZ$, where $a\in{\mathbb{R}},$ has a normal distribution with parameters $a\mu_Z,a^2\sigma^2_Z$. We receive, that $\frac{X_2}{2}$ has a normal distribution with parameters $\frac{\mu_2}{2},\frac{\sigma^2_2}{4}$ and $\frac{X_3}{3}$ has a normal distribution with parameters: $\frac{\mu_3}{3},\frac{\sigma^2_3}{9}$. We use also a known fact, that a sum of normally distributed variables is again a normally distributed variable. We get, that $Y_2$ has a normal distribution with parameters: $\mu_1+\frac{\mu_2}2,\sigma_1^2+\frac{\sigma_2^2}{4}$ and $Y_3$ has a normal distribution with parameters: $\mu_1+\mu_2+\frac{\mu_3}3,\sigma_1^2+\sigma_2^2+\frac{\sigma_3^2}{9}$. Denote: $\alpha_1=\mu_1+\frac{\mu_2}2$, $\beta_1=\sqrt{\sigma_1^2+\frac{\sigma_2^2}{4}}$, $\alpha_2=\mu_1+\mu_2+\frac{\mu_3}3$, $\beta_2=\sqrt{\sigma_1^2+\sigma_2^2+\frac{\sigma_3^2}{9}}$. We use a known fact that the joint probability density function is equal to the product of density functions for independent variables. We get: $p_{Y_1,Y_2,Y_3}(y_1,y_2,y_3)=p_{Y_1}(y_1)p_{Y_2}(y_2)p_{Y_3}(y_3)=\frac{1}{\sigma_2\beta_1\beta_2(\sqrt{2\pi})^3}e^{-\frac12(\frac{y_1-\mu_2}{\sigma_2})^2-\frac12(\frac{y_2-\alpha_1}{\beta_1})^2-\frac12(\frac{y_3-\alpha_2}{\beta_2})^2}$

Answer: $p_{Y_1,Y_2,Y_3}(y_1,y_2,y_3)=\frac{1}{\sigma_2\beta_1\beta_2(\sqrt{2\pi})^3}e^{-\frac12(\frac{y_1-\mu_2}{\sigma_2})^2-\frac12(\frac{y_2-\alpha_1}{\beta_1})^2-\frac12(\frac{y_3-\alpha_2}{\beta_2})^2}.$