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The funnel distribution for random variable $X = (x_1,x_2,..,x_D)$is

$$P(X) = N(x_1|0,9)\prod_{d=2}^D N(x_d | 0,exp(x_1))$$

The closed form normalizing constant for normal distribution $N(x|0,\sigma^2)$ is $\frac{1}{\sqrt{2\pi\sigma^2}}$.

However, in this case, the variance of $x_2, ..x_d$ is depended on $x_1$.

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Let $X'=(x_2,x_3,...,x_D)$, then $P(X)=P(X'\mid x_1)P(x_1)$.

$\int_{-\infty}^{\infty} N(x_1|0,9) (\idotsint_{D-1} (\prod_{d=2}^D N(x_d | 0,exp(x_1))) dx_2 \dots dx_{D})dx_1= \int_{-\infty}^{\infty} N(x_1|0,9) (\sqrt{2\pi exp(-x_1)})^{D-1}dx_1= (\sqrt{2\pi})^{D-1} \int_{-\infty}^{\infty} exp(-\frac{x_1^2+9(D-1)x_1}{18})=(\sqrt{2\pi})^{D}exp(\frac{9(D-1)^2}{8}).$

Problem reduces to finding normalizing constant for single integration. Just need to complete the square.

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  • $\begingroup$ Thanks for the reply; but I'm still confused that how does the second equation gives the normalizing constant ? $\endgroup$ – ElleryL Dec 9 '18 at 1:10
  • $\begingroup$ Answer updated. Hope this makes sense! $\endgroup$ – user128949 Dec 9 '18 at 1:39

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