If $W_1=Y_1+2Y_2$ and $W_2=4Y_1-Y_2$ what is the joint distribution of $W_1$ and $W_2$

Question

Let $Y_1$ and $Y_2$ be independent random variables with $Y_1\sim N(1,3)$ and $Y_2 \sim N(2,5).$

If $W_1=Y_1+2Y_2$ and $W_2=4Y_1-Y_2$ what is the joint distribution of $W_1$ and $W_2$?

Is correct my procedure?

$E(W_1)=E(Y_1+2Y_2)=E(Y_1)+2E(Y_2)=1+2(2)=5$

$Var(W_1)=var(Y_1+2Y_2)=var(Y_1)+4var(Y_2)=3+4(5)=23$

$E(W_2)=E(4Y_1-Y_2)=4E(Y_1)-E(Y_2)=4(1)-2$

$Var(W_2)=var(4Y_1-Y_2)=16var(Y_1)-var(Y_2)=16(3)-5=43$

$f(w_1)=\frac{1}{\sqrt{2\pi}23}\epsilon^{\frac{-1}{2}(\frac{x-5}{23})^2}$

$f(w_2)=\frac{1}{\sqrt{2\pi}43}\epsilon^{\frac{-1}{2}(\frac{x-2}{43})^2}$

$f(w_1,w_2)=\frac{1}{\sqrt{2\pi}23}\epsilon^{\frac{-1}{2}(\frac{x-5}{23})^2} \cdot \frac{1}{\sqrt{2\pi}43}\epsilon^{\frac{-1}{2}(\frac{x-2}{43})^2} $

Answer 1

"Mostly" right, good progress. But you aren't quite finished.

Because $W_1$ and $W_2$ are both influenced by the same $Y_1$ and $Y_2$, it seems intuitively clear that you need to find the covariance also. Remember that $Cov$ is linear in both its arguments and that the $Y_i$ are independent.

Here is a brief simulation with answers that should be accurate to 2 (maybe 3) places. Answers are based on a million performances of the experiment to sample $Y_1$ and $Y_2$ and transform them to $W_1$ and $W_2.$ You can check the answers you have (I see an error) and then the covariance that you need to find. Finally, you need to look at the formula for the bivariate normal density function.

 m = 10^6;  y1 = rnorm(m, 1, sqrt(3));  y2 = rnorm(m, 2, sqrt(5))
 w1 = y1 + 2*y2;  w2 = 4*y1 - y2
 mean(y1);  sd(y1);  mean(y2);  sd(y2);  cov(y1,y2)
 ## 0.9951616
 ## 1.732375
 ## 2.001808
 ## 2.235283
 ## -0.003585973  # Consistent with indep y1 and y2

 mean(w1);  sd(w1);  mean(w2);  sd(w2)
 ## 4.998779
 ## 4.792989
 ## 1.978838
 ## 7.283074
 sqrt(23)
 ## 4.795832
 sqrt(43)      
 ## 6.557439  # Better check formula for variance of difference

 cov(w1,y2); cor(w1, w2)
 ## 9.989395
 ## 0.05690479

Even if someone is eager to show off the right answers, you will learn something by looking at these approximate answers, and reviewing the necessary relationships. Please leave me a Comment if there is something you can't figure out.