Estimating a Probability Density Function for the Mean of a Normal Distribution

Question

I have been presented with the following problem:

1. I don't know the mean of a Gaussian random variable;
2. Will consider this mean as another random variable (M). M can take the values $Z_1$ with probability $p$ and $Z_2$ with probability $(1-p)$, where $p ∈ (0, 1)$.
3. I know that $Z_1 \sim N(μ_1, σ_1)$ and $Z_2 \sim N(μ_2, σ_2)$.
4. What is the density of M?

By the Law of Total Probability I got to this expression: $$ P(M) = P(M|Z_1)P(Z_1) + P(M|Z_2)P(Z_2) $$

Can I consider $P(M)$, as presented above, as the density of M?

And, does $P(M|Z_1)$ represent the density of the Normal distribution with mean $μ_1$ and variance $σ_1$?

I am afraid I am mixing some statistical concepts here, I really appreciate if you could enlighten me, thank you!

Answer 1

First of all, $P(M)$ is not a density it's a probability.

With 2. we have $M = p*Z_1 + (1-p)*Z_2$ where $Z_1$ and $Z_2$ are Gaussian distributions.

So we can say that $M$ is a linear combination of two normal distribution.

This gives us directly the distribution of $M$ which is going to be : $$M∼N(p*μ_1+(1-p)*μ_2, p^2*σ_1^2+(1-p)^2*σ_2^2)$$

Now it's easy to have the density of $M$ which is going to be: $$f_M(m) = \frac{1}{\sqrt(2*\pi)σ_m}* \exp(-\frac{(m-μ_m)^2}{2σ_m^2})$$ where we have :

$μ_m = p*μ_1+(1-p)*μ_2 $
$σ_m^2 = p^2*σ_1^2+(1-p)^2*σ_2^2$

EDIT: $Z_1$ and $Z_2$ should be independent. $Z_1∼N(μ_1,σ_1^2)$ and $Z_2∼N(μ_2,σ_2^2)$