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$$\Sigma_{1,2}=\Sigma_{1,1}-\Sigma_{1,2}\Sigma_{22}^{-1}\Sigma_{2,1}$$ $$\mu_{1,2}=\mu_1+\Sigma_{1,2}\Sigma_{2,2}^{-1}(x_2-\mu_2)$$ This indicates that conditional variance of multivariate normal distribution is independent of $x_2$ but the conditional mean is dependent of $x_2$

https://en.wikipedia.org/wiki/Multivariate_normal_distribution

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However, intuitively, no matter where $x_2$ you slice, you always get the same mean. But if you take the slice closer to the origin the variance is bigger (flatter normal curve on another direction) and smaller if the slice is further to the origin. Isn't this counter intuitive?

laurab
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ZHU
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  • You mistake the width of the cross-section with the variance. When examining a cross-section of the curve, you must rescale the heights to integrate to unity. With a Bivariate Normal distribution when you do that, all the cross sections have identical variances. For more intuition, refer to the analysis and graphics at https://stats.stackexchange.com/a/71303/919. – whuber Aug 19 '23 at 15:00

1 Answers1

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Your question is somewhat unclear, you say

However intuitively no matter where $x_2$ you slice you always get the same mean while if you take the slice closer to the origin the variance is bigger (flatter normal curve on another direction) and smaller if the slice is further to the origin.

I don't understand why you say you get the same mean ... Note that if the bivariate density is $f(x, y)$ then the conditional density is $$ f(x \mid y) = \frac{f(x, y)}{f(y)} $$ that is, visually you find it by slicing the bivariate density at the given value of $y$, and the rescaling by division with $f(y)$ to secure the conditional density integrates to 1. I guess that by looking only at the bivariate density, and doing the slicing mentally, you forget the rescaling, and gets the wrong impression.

Some Details

I do not know about the intuition, but let us look in more detail at the normal bivariate case. To simplify the algebra we assume that $$ \mu_X = \mu_Y = 1; \sigma_X=\sigma_Y=1 $$ Then calculate the conditional density of $X$ given $Y=y$ (details is an exercise). Write down this conditional density of $X$ for $Y=y_1$ and then for $Y=y_2$, and calculate $$ \frac{f(x \mid Y=y_1)}{f(x \mid Y=y_2)} = \exp\left\{ -\frac{\rho}{1-\rho^2}x (y_1- y_2) - \frac12 \frac{\rho^2}{1-\rho^2} (y_1^2 - y_2^2 ) \right\} $$ With other worlds, the conditional densities of $x$ for different values of $y$, have the same shape.

kjetil b halvorsen
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