Let X1 and X2 two independent normal distributions. Compute the joint distribution of (X1+X2, X1+2X2)

i am faced with the following problem which i can't seem to solve...

Let $$X_1$$ and $$X_2$$ two independent normal distributions of parameters $$\mu$$ and $$\sigma^2$$ and let $$Y_1 = X_1 + X_2$$ and $$Y_2 = X_1 + 2X_2$$.

Compute the joint probability distribution of $$Y_1$$ and $$Y_2$$

This is what i tried

$$f_{Y_2\vert Y_1}(y_2)f_{Y_1}(y_1) = f_{Y_1,Y_2}(y_1, y_2)$$ For a given $$x_1 + x_2 = y_1$$, we have $$x_1 + 2x_2 = y_2 \iff x_2 = y_2 - y_1$$

Hence $$f_{Y_2\vert Y_1}(y_2) = f_{X_2}(y_2 - y_1)$$ $$f_{Y_1,Y_2}(y_1, y_2) = f_{X_2}(y_2 - y_1)f_{Y_1}(y_1)$$

$$Y_1 = X_1 + X_2 \implies Y_1$$ is normal of parameters $$2\mu$$ and $$2\sigma^2$$ and $$X_2$$ is normal of parameters $$\mu$$ and $$\sigma^2$$ so we can easily compute the product.

However, with this approach, i do not get the desired result... Could anyone see where this reasoning fails?

The Jacobian transformation you attempted should go:

$\left\{\raise{1ex}{Y_1=X_1+X_2\\Y_2=X_1+2X_2}\right\} \iff\left\{\raise{1ex}{X_1=2Y_1-Y_2\\X_2=Y_2-Y_1}\right\}$

So therefore $f_{Y_1,Y_2}(u, v) ~=~ \begin{Vmatrix}\dfrac{\partial (2u-v,v-u)}{\partial (u,v)}\end{Vmatrix} f_{X_1}(2u-v)f_{X_2}(v-u)$

• Thanks! This is what i was looking for! Sep 18 '17 at 0:37

One should say "Let $X_1$ and $X_2$ two independent normally distributed random variables", i.e. it is the random variables, not the distributions, that are independent.

One has $\operatorname{E}(Y_1) = 2\mu$ and $\operatorname{E}(Y_2) = 3\mu,$ and then \begin{align} \operatorname{var}(Y_1) & = 2\sigma^2, \\[10pt] \operatorname{var}(Y_2) & = \operatorname{var}(X_1) + 2^2 \operatorname{var}(X_2) = 5\sigma^2, \\[10pt] \operatorname{cov}(Y_1,Y_2) & = \operatorname{cov}(X_1+X_2, X_1+2X_2) \\ & = \operatorname{cov}(X_1,X_1+2X_2) + \operatorname{cov}(X_2,X_1+2X_2) \\ & = \Big( \operatorname{cov}(X_1,X_1)+2\operatorname{cov}(X_1,2X_2)\Big) + \Big(\operatorname{cov}(X_2,X_1)+ \operatorname{cov}(X_2,2X_2) \Big) \\ & = \sigma^2 + 0 + 0 + 2\sigma^2 = 3\sigma^2. \end{align}

One can say that the vector $\begin{bmatrix} Y_1 \\ Y_2 \end{bmatrix}$ has a bivariate normal distribution with expected value $\begin{bmatrix} 2\mu \\ 3\mu \end{bmatrix}$ and variance $\begin{bmatrix} 2\sigma^2 & 3\sigma^2 \\ 3\sigma^2 & 5\sigma^2 \end{bmatrix}.$

That is one way of understanding the question of what the distribution is.

• I think you have the wrong expected value in the sentence "One can say..." Sep 17 '17 at 22:47
• @Michael Thank you for the answer! However, we haven't worked with bivariate normal distributions nor variance-covariance matrices, and i think i am supposed to solve this only using conditioning and joint distributions. Do you have any idea how one could work with those tools only? Sep 17 '17 at 22:56