# Computing a marginal distribution of a joint involving a delta function

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 $$y$$.

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: $$h(x_0)=f(v,z|x_0)$$