# Prove ($Y_1$,$Y_2$) is normal bivariate if $Y_1$ = $X_1$ and $Y_2$=$X_1 + X_2$ with $X_i \backsim N(\mu_i,\sigma_i^2)$ independent

I'm struggling a little with this question. It's clear to me that $$Y_2$$ is normal because it is a linear combination of two normals. However, as I understand it, not EVERY vector of univariate normals is normal, and so proving ($$Y_1$$,$$Y_2$$) is bivariate is not as trivial as stating that its components are normals.

I have tried to do so by finding $$f(y_1,y_2)$$ from $$f(y_1)$$ (which is a normal) and $$f(y_2|y_1)$$ (which can be found using the conditional cdf of $$y_2$$ given $$y1$$). However, the algebra doesn't quite work out and I'm missing a few terms. I also tried it using the integral transformation theorem from ($$X_1,X_2$$) to ($$X_1,X_1+X_2$$) but I get the same result.

I currently have this: with $$\mu_1 = E[X_1]$$, $$\sigma_1^2 = Var(X_1)$$, $$\mu_2 = E[X_2]$$, $$\sigma_2^2 = Var(X_2)$$

$$f(y_1,y_2) = \frac{1}{\sqrt{2\pi}\sigma_1}e^{\frac{-1}{2}(\frac{y_1-\mu_1}{\sigma_1})^2} * \frac{1}{2\sqrt{\pi}\sigma_2}e^{\frac{-1}{2}(\frac{y_2-y_1-\mu_2}{\sigma_2})^2}$$

I got this through both methods, but I have no idea how to get the normal bivariate density function from there.

¿Is there another way to prove that the vector is bivariate normal? ¿Or any hints with the algebra required.

• I think you should mention somewhere that $X_1$ and $X_2$ are independent for your conclusion to hold
– Momo
Commented Nov 2, 2021 at 21:09

Hint: $$Y_1$$ and $$Y_2$$ are correlated, so you need to write your pdf in the general Bivariate Normal probability density function (re-written below for the record):

$$f(y_1,y_2)=\frac{1}{2\pi\sigma_{y_1}\sigma_{y_2}\sqrt{1-\rho^2}}\exp\left[-\frac{z}{2(1-\rho^2)}\right]$$ with: $$z=\frac{(y_1-\mu_{y_1})^2}{\sigma_{y_1}^2}-\frac{2(y_1-\mu_{y_1})(y_2-\mu_{y_2})}{\sigma_{y_1}\sigma_{y_2}}+\frac{(y_2-\mu_{y_2})^2}{\sigma_{y_2}^2}$$ and: $$\rho=\frac{\operatorname{Cov}(Y_1,Y_2)}{\sigma_{y_1}\sigma_{y_2}}$$

In your case you have $$\mu_{y_1}=\mu_1,\ \mu_{y_2}=\mu_1+\mu_2,\ \sigma_{y_1}^2=\sigma_1^2,\ \sigma_{y_2}^2=\sigma_1^2+\sigma_2^2,\text{ and }\rho=\frac{\sigma_1}{\sqrt{\sigma_1^2+\sigma_2^2}}$$

If you know matrix algebra and you know that non-singular linear transformations of multivariate normal random variables is also multivariate normal, you can write:

$$\begin{bmatrix} Y_1 \\ Y_2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \end{bmatrix}$$

$$\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$$ is nonsingular, $$\begin{bmatrix} X_1 \\ X_2 \end{bmatrix}$$ is multivariate normal, and voila!

• Yes! I got that far (calculating the correlation correctly). But it's exactly that what's bothering me. How would I get $\sqrt{1-\rho^2}$ into the denominator if it doesn't show up in either (and multiplying both numerator and denominator si weird since then I would have it in the numerator where it doesn't go Commented Nov 2, 2021 at 20:39
• Do you mean $\sigma_{y1}\sigma_{y2}\sqrt{1-\rho^2}=\sigma_1\sqrt{\sigma_1^2+\sigma_2^2}\sqrt{1-\frac{\sigma_1^2}{\sigma_1^2+\sigma_2^2}}=\sigma_1\sigma_2$? BTW, in your formula for $f(y_1,y_2)$ I think that $2$ should be inside the square root.
– Momo
Commented Nov 2, 2021 at 21:01
• The linear algebra solution was exactly what I needed. Thanks! Commented Nov 3, 2021 at 2:19