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I am a newbie to econometrics, so kindly excuse me if I sound too naive.

This is what Fumio Hayashi says on page 34 of "Econometrics":

Recall from probability theory that the normal distribution has several convenient features:

• The distribution depends only on the mean and the variance. Thus, once the mean and the variance are known, you can write down the density function. If the distribution conditional on X is normal, the mean and the variance can depend on X. It follows that, if the distribution conditional on X is normal and if neither the conditional mean nor the conditional variance depends on X, then the marginal (i.e., unconditional) distribution is the same normal distribution. (Italics my own).

I am riddled with the italicised part. In ${{\mathit f}({\mathbf z} {\mid} {\mathbf x)}}$ = ${\frac{\mathit f({\mathbf z},{\mathbf x})}{\mathit f({\mathbf x})}}$, if this conditional distribution is normal (${\mathcal N} ({\mu}, {\Sigma})$), in which case, the only parameters that affect the conditional density are its mean ($\mu$) and variance ($\Sigma$), X is still present in the density function, and hence the distribution does depend on X (Am I right in my understanding of conditional normality, or am I missing something?).

So to rephrase, my concern is how can ${{\mathit f}({\mathbf z} {\mid} {\mathbf x)}}$ be the same as the unconditional distribution ${\mathit f}$($\mathbf {z}$), when the former is dependent on X, even if it were to be the case that ${\mu}$ and ${\Sigma}$ do not depend on X.

Many Thanks for the Help...

ekss.

gunes
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eisendon
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1 Answers1

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It says if neither the mean nor the variance depends on $X$ the conditional distribution is equal to the unconditional one. Because if $\mu,\Sigma$ were constant vectors/matrices, then there is no other parameter left in the normal distribution you can fit $X$ into. It implicitly says that $f(z|x)$ doesn't have any $x$ in it, e.g. a simpler example from exponential distribution:

$$f(z|x)=\lambda e^{-\lambda z}$$ and $\lambda$ doesn't depend on $x$. It means $f(z|x)=f(z)\ \ \forall x$.

gunes
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  • Thanks a lot, but one last point :- doesn't X figure (either directly, or in a functional form along with z), in the random variable for which the conditional density is being examined, which undertakes normality in this case? If not, how do we know this before hand, since we do not have prior knowledge that X & z are independent. How are we to know this from and Σ? – eisendon Apr 02 '20 at 11:19
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    In your first sentence, do you mean $z$ can actually be a function of $x$, i.e. $z(x)$? If it is, then no. $z$ is another variable in the joint. And, you never know it beforehand unless you know the joint distribution. In this case, you have it. – gunes Apr 02 '20 at 11:29
  • Actually, my intention in that sentence was to ask whether X could be present in *any* form in ${\mathit f}$ (z|x). Because if that is so, then the conditional density does depend on X. However, if this is not so, then I will have to go back to my books. I would very well appreciate any hints from your side though, regarding the logic. – eisendon Apr 02 '20 at 11:40
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    Yes, X can't be present in any form in $f(z|x)$ if z and x are independent. But, if dependent, it'll be present somehow, e.g. $$f(z|x)=x^2e^{-x^2z}$$ In your question, the form of the conditional density is assumed to be normal, so if none of the mean and the covariance matrix don't depend on $x$, the conditional density won't depend on $x$. – gunes Apr 02 '20 at 11:45
  • How to prove: "if mean and variance of $\mathit {f}$ (z|x) don't depend on x, the conditional density won't depend on x"? If you can edit your answer to make space for this proof, I'd be thrilled. Thanks – eisendon Apr 02 '20 at 13:51
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    doesn't "there is no other parameter left in the normal distribution you can fit into" answer it? In a density you have the primal variable, $z$, the parameters of the density, and constants. The primal variable is different then the conditioned var, as the name suggests the constants cannot depend on the conditioned var, and if also the parameters in the density don't depend on the conditioned var, what else left? – gunes Apr 02 '20 at 14:12