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What are the conditions under which a distribution reduces down a dimension?

For example, suppose I have a 2D gaussian distribution for X and Y. Under what condition(s) on Y does the distribution "reduce" down to a 1D gaussian distribution for X?

I don't want to say more so as not to bias the answers.

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    Constant $Y$ seems to meet the condition if you allow a constant to be described as Gaussian. So too does a bijective relationship between $X$ and $Y$ (i.e. knowledge of $X$ determines the value of $Y$ with probability 1) which preserves the Gaussian properties.2011-04-04
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    A constant Y would have a sigma of exactly zero. I can't figure out how to to reduce 2D gaussian to a 1D guassian in this case. I keep trying to use a limit as sigma goes to zero but it doesn't work right. I don't want to assume a relationship between X and Y.2011-04-04
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    a constant $Y$ would have a $\sigma \to \infty$...2011-04-04
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    I'm sorry. I don't follow that, Fabian. If Y is constant. Every time I measure it, I will get Y. The dispersion is therefore zero. Could you explain further?2011-04-04
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    I'm stumbled across the central limit theorem, which may contain the answer to my question. We'll see.2011-04-04
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    @Henry Isn't the _only_ bijective relationship between $X$ and $Y$ such that both $X$ and $Y$ are Gaussian a linear (or affine) relationship: $Y = aX+b$ for real numbers $a$ and $b$ with $a$ being allowed to be $0$ if we are amenable to calling a constant a Gaussian random variable with zero variance?2011-10-30

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Given your assumptions, the variance of Y should be much less than the variance of X. More generally, when the "principal component" is neither X nor Y but a combination of the two, the solution is given by Principal component analysis. More generally still, see Dimension reduction.

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I don't think there are any such conditions, unless $\Sigma$ is allowed to be non-negative definite (and not strictly positive definite). If either of the marginal variances approaches 0 or the correlation between the two approaches 1, the distribution effectively collapses to 1 dimension and you still have a non-negative definite $\Sigma$. However, I don't think this is legal because $\Sigma$ would be singular and the joint density would be undefined.