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Suppose $X_1$ and $X_2$ are bivariate normal with mean $\mu=(\mu_1,\mu_2)$ and covariance $\Sigma = \begin{bmatrix} \sigma_{11} & \sigma_{12} \\ \sigma_{12} & \sigma_{22} \\ \end{bmatrix}$. What is the probability density function $f\left(X_1=x_1,X_2=x_2|X_1-X_2 = w\right)$?

Answer attempt

Using the definition of conditional probability \begin{align} f\left(X_1=x_1,X_2=x_2|X_1-X_2 = w\right) &= \frac{f\left(X_1=x_1,X_2=x_2,X_1-X_2 = w\right)}{f\left(X_1-X_2 = w\right)}\\ &= \frac{f\left(X_1=w-x_2,X_2=x_2\right)}{f\left(X_1-X_2 = w\right)}\\ \end{align}

The pdf of $f\left(X_1=w-x_2,X_2=x_2\right)$ is the bivariate normal with parameters $\mu$ and $\Sigma$. The pdf of $f\left(X_1-X_2 = w\right)$ is univariate normal with mean $\mu_1-\mu_2$ and variance $\sigma_{11}+\sigma_{22} - 2\sigma_{12}$.

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    $\begingroup$ First, check your algebra--it's easy to get signs wrong. Then, consider why $w$ doesn't appear anywhere in your answer. Finally, ask yourself whether your answer would be of much use. Most likely the intended answer for this exercise will be expressed as a function of $\mu_1,\mu_2,\sigma_{ij},$ and $w.$ Since it evidently is a part of a study of the bivariate Normal distribution, you probably have previously studied regression--so why not apply that and simply regress $(X_1,X_2)$ against $X_1-X_2$? $\endgroup$ – whuber Jul 30 '18 at 14:57

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