How to get the conditional probability of $X|Y\sim Y N(0,\sigma_1^2)+(1-Y) N(0,\sigma_2^2)$?

Question

If we have a random variable $Y\sim Ber(p)$ and $$X|Y\sim Y N(0,\sigma_1^2)+(1-Y)N(0,\sigma_2^2)$$

How to get $f(x|y)$ proportion to the product of two normal distribution?

Is it just the sum of two normal distributions? The expectation of $X|Y$ is $\mathbb{E}(X|Y)=0$. Do we have $ Var(X|Y)=Y^2\sigma_1^2+(1-Y)^2\sigma_2^2$

Answer 1

Setting $Z_i\sim N(0;\sigma_i^2)$ and assuming independence between $Z_1$,$Z_2$ and $Y$,

your conditional density is a mixture of the two gaussian densities

$$f_{X|Y=y}(t)=yf_{Z_1}(t) +(1-y)f_{Z_2}(t)$$

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The marginal distribution is:

$$f_X(x)=\frac{p}{\sigma_1\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma_1^2}}+\frac{1-p}{\sigma_2\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma_2^2}}$$

Answer 2

The question is not 100% clear so I will get everything in one place.

Conditional distribution of $X|Y$ is normal whose parameter depends on $Y$, with density function $$ f_{X|Y}(x|y) = y \frac{1}{\sqrt{2\pi}\sigma_1}\exp\left(-\frac{x^2}{2\sigma_1^2}\right) + (1-y) \frac{1}{\sqrt{2\pi}\sigma_2}\exp\left(-\frac{x^2}{2\sigma_2^2}\right), x \in (-\infty, \infty), y \in \{0,1\} $$

Marginal distribution of $X$ is a mixture distribution with density $$f_X(x) = p \frac{1}{\sqrt{2\pi}\sigma_1}\exp\left(-\frac{x^2}{2\sigma_1^2}\right) + (1-p) \frac{1}{\sqrt{2\pi}\sigma_2}\exp\left(-\frac{x^2}{2\sigma_2^2}\right), x \in (-\infty, \infty)$$

Conditional variance of $X|Y$is a function of $Y$: $$ \text{var}(X|Y) = Y\sigma_1^2 + (1-Y) \sigma_2^2 $$ Marginal variance of $X$ is a number: $$ \text{var}(X) = p\sigma_1^2 + (1-p) \sigma_2^2. $$