# Distribution of $(X_1,X_2)$ if $X_1\pm X_2$ are two independent $N(1,4)$

$$X_1+X_2$$ and $$X_1-X_2$$ are i.i.d. $$N(1,4)$$. What is the distribution of $$X = (X_1,X_2)^T$$?

I know i.i.d. is an independent and identically distributed random variable but I don't know how to use it to solve this problem that contains a column vector.

• What does this have to do with computer science? Apr 4 at 20:19
• Are you familiar with multivariate normal distributions? Apr 4 at 20:27
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– D.W.
Apr 5 at 4:50

## 1 Answer

Let $$Y = (Y_1,Y_2)$$ be two i.i.d. $$N(1,4)$$ variables. Then $$Y$$ is a multivariate normal distribution, and so any affine transformation of $$Y$$ is a multivariate normal distribution. In particular, if we define $$X_1 = \frac{Y_1 + Y_2}{2}$$ and $$X_2 = \frac{Y_1 - Y_2}{2}$$ then $$X = (X_1,X_2)$$ is a multivariate normal distribution which satisfies $$X_1 + X_2 = Y_1$$ and $$X_1 - X_2 = Y_2$$. In other words, it is the random variable you're interested in.

It is routine to calculate $$(X_1,X_2) \sim N\left(\begin{pmatrix} 1 & 0 \end{pmatrix}, \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\right),$$ that is, $$X_1 \sim N(1,2)$$ and $$X_2 \sim N(0,2)$$, and $$X_1,X_2$$ are independent.