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I'd like to ask for advice tackling a new (to me) problem in conditional distributions.

Let $X$ and $Y$ be independent real-valued random vectors. Define $Z = H(X) + Y$ where $H$ is a nonlinear, possibly non-invertable map. I want to compute $f_X(x|Z=z^*)$: the posterior distribution of $X$ given the datum $Z = z^*$ (that is, I can measure only $Z$, not $X$ or $Y$, and my measurement of $Z$ is $z^*$) while assuming a prior distribution $f_X(x)$ of $X$ and a known distribution $f_Y(y)$ of $Y$.

My attempt is as follows:

From Bayes' theorem we have

$$f_{X}(x|Z=z^*) = \frac{f_{Z}(Z=z^*|X=x)f_X(x)}{\int f_Z(Z=z^*|X=x')f_X(x')dx'},$$

So we really need $f_{Z}(Z=z^*|X=x)$. I interpreted this as the distribution of $Z$ assuming $X$ is distributed according to $f_X(x)$, which is then evaluated at the specific point $z^*$ (I know there is some subtlety about sets of measure zero here and I may have to take a limit, but I'm not sure of the best way to do this).

Suppose I can determine the distribution of the variable $\phi = H(X)$ from knowledge of $f_X(x)$ (which I know is possible using the Jacobian and inverse map in the case where $H$ is 1-1, for instance). Let's label the distribution of $\phi$ if $X\sim f_X(x)$ by $\pi(\phi)$--then, since $X$ and $Y$, thus $\phi$ and $Y$, are independent, Theorem 7.1 of this PDF implies $f_Z(z)$ is a convolution of $\pi$ and $f_Y$. I get

$$ f_{Z}(Z=z^*|X=x) = \int \pi(z^*-y)f_Y(y)dy $$

for the needed conditional distribution and thus

$$ f_{X}(x|Z=z^*) = \frac{f_X(x)\int \pi(z^*-y)f_Y(y)dy}{\int \int \pi(z^*-y)f_Y(y)f_X(x')dx'dy}, $$

for the final result.

My questions are as follows:

  1. Have I interpreted the conditional $f(Z|X=x)$ correctly? Is it just the distribution of $Z(X,Y)$ assuming $X\sim f_X(x)$? (I am concerned it is more complicated than this).

  2. Is my intuition of evaluating $f_Z(z|X=x)$ at $z^*$ as the way to account for the fact that the measurement of $Z$ was $z^*$ correct? (Same concern as above).

  3. Is there a general formula for the distribution of $\phi$ given $X \sim f_X(x)$ even when $H$ is not invertible? What about the case where $H$ is simply a linear mapping: $H(X) = Hx$?

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Theorem 7.1 concerns the distribution of a sum of two random (discrete) variables. That is not what you are looking for, which is the conditional distribution of the sum when given one of the variables.

Since $H()$ is fully determined for any given value of $X$, and as $X,Y$ are independent, thus:

$$f_{Z\mid X}(z\mid x) {= f_{H(X)+Y\mid X}(z\mid x)\\ = f_{Y\mid X}(z-H(x)\mid x)\\ = f_{Y}(z-H(x))} \\~\\ f_{Z,X}(z,x) = f_Y(z-H(x))\; f_X(x)$$

Notice the last line is the joint distribution of $Z$ and $X$.   Obtaining the marginal for $Z$ from this gives the convolution.

So therefore, using this.

$$f_{X\mid Z}(x\mid z^*)~{=\dfrac{f_{Z\mid X}(z^*\mid x) f_X(x)}{\int_\Bbb R f_{Z\mid X}(z^*\mid s) f_X(s)\;\mathrm d s} \\[4ex] = \dfrac{f_Y(z^*-H(x))\;f_X(x)}{\int_\Bbb R f_Y(z^*-H(s))\;f_X(s)\operatorname d s } }$$

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  • $\begingroup$ Thanks! Don't know how I missed that! $\endgroup$ – SZN Aug 21 '17 at 3:45

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