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Suppose that we have four continuous random variables $x,y,z,$ and $v$ and we want to compute the following integral:

$$\int f(x\mid y)f(z\mid x,y)f(v\mid z,x,y)\,dx$$

There are a few conditions:

  1. $f(z\mid x,y) = f(z\mid x)$
  2. $f(v\mid x,y,z) = f(v\mid z,x)$
  3. $f(x\mid y) = \delta_x(x_0\mid y)$

where $\delta_x(x_0\mid y)$ is a Dirac delta function where the value of $x_0$ depends on $y$. Is the following calculation of this integral correct?

$$\int \delta_x(x_0\mid y)f(z\mid x)f(v\mid z,x)\,dx = f(z,v\mid x_0(y))$$

I used $x_0(y)$ since $x_0$ is a function of $y$.

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I presume you mean $\delta(x-x_0(y))$ by $\delta_x(x_0(y))$. By Translation property, we have (if $x_0\in\mathcal{X}$): $$\int_\mathcal{X} h(x)\delta(x-x_0)dx=h(x_0)$$ The expression in the integral is a function of $x$, since $v,z$ are regarded as constants in the integral,so we have $h(x)=f(z|x)f(v|z,x)=f(v,z|x)$. Therefore, the result is $h(x_0)$, as in yours: $$h(x_0)=f(v,z|x_0)$$

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