Entropy on a draw from a random distribution. Suppose I am attempting to calculate the entropy of a continuous, normally distributed random variable $X$, from the distribution $\mathcal{N}(\mu, \sigma)$. This is easy to to do - I just calculate 
$$
h[f] = -\int_{\mathbb{R}} f(x) \ln(f(x)) dx.
$$
Here $f(x)$ is the pdf of a normal variable $\mathcal{N}(\mu, \sigma)$. 
Now, suppose I have a more complicated random variable, $Y$, which is this: first draw random values $\mu_i$ and $\sigma_i$ uniformly from a pre defined set of values $\{\mu_k\}_{k=1}^M$ and $\{\sigma_k\}_{k=1}^M$, and then draw $Y$ from the distribution $\mathcal{N}(\mu_i, \sigma_i)$.What is the entropy of the random variable, $Y$? 
I would love if the answer to this question involved any of the following terms:
$$
\mathbb{E}(\{\sigma_k\}_{k=1}^M), \mathbb{E}(\{\mu_k\}_{k=1}^M), \mathbb{V}(\{\sigma_k\}_{k=1}^M), \mbox{ or } \mathbb{V}(\{\mu_k\}_{k=1}^M),
$$
although any and all help is appreciated. ($\mathbb{E}$ and $\mathbb{V}$ are expectation and variance, respectively.)
More general answers are of course welcome as well!
 A: *

*To find the entropy of $Y$, note that we have 
$$\Pr(Y\le y)=\sum_{i,j}\Pr(Y\le y\mid \mu_i,\sigma_j)\Pr(\mu_i,\sigma_j)$$
and so
$$f_Y(y)=\sum_{i,j}f_{N(\mu_i,\sigma_j)}(y)\Pr(\mu_i,\sigma_j)$$
which can be substituted into
$$h[f_Y]=-\int f_Y(x)\ln(f_Y(x))dx.$$

*Regarding this part,

"I would love if the answer to this question involved any of the following terms:
  $$
\mathbb{E}(\{\sigma_k\}_{k=1}^M), \mathbb{E}(\{\mu_k\}_{k=1}^M), \mathbb{V}(\{\sigma_k\}_{k=1}^M), \mbox{ or } \mathbb{V}(\{\mu_k\}_{k=1}^M),"
$$

I'm guessing it's not possible. Following an example by @AndreasBlass, choose any real number $a>1$ and consider the distribution for $\mu_i$ giving the points $-a,0,a$ probabilities $p,1-2p,p$ where $p=1/2a^2$. This has mean $1$ and variance $1$ for all $a$. The entropy of this distribution is $-2p\ln(p)-(1-2p)\ln(1-2p)$ which does depend on $p$.
Now let us say $\sigma$ must be constant 1, but $\mu$ follows the distribution just mentioned. Then the entropy of $\mu$ is not determined by the mean and variance of $\mu$. You could, I suppose, check whether the entropy of $Y$ depends on $a$.

For what it's worth, the pdf of $Y$ in this case is
$$f_Y(y)=f_{N(-a,1)}(y)p + f_{N(0,1)}(y)(1-2p) + f_{N(a,1)}(y)p$$
$$=f_{N(-a,1)}(y)\frac{1}{2a^2} + f_{N(0,1)}(y)(1-\frac{1}{a^2}) + f_{N(a,1)}(y)\frac{1}{2a^2}$$
$$=f_{N(-a,1)}(y)\frac{1}{2a^2} + f_{N(0,1)}(y)(1-\frac{1}{a^2}) + f_{N(a,1)}(y)\frac{1}{2a^2}$$
This becomes (multiplying out the $\sqrt{2\pi}$)
$$
e^{-(y-(-a))^2/2}\frac{1}{2a^2} + e^{-y^2/2}\left(1-\frac{1}{a^2}\right) + e^{-(y-a)^2/2}\frac{1}{2a^2}
$$
$$
=\left(e^{-(y-a)^2/2} + e^{-(y-(-a))^2/2}\right)\frac{1}{2a^2} + e^{-y^2/2}\left(1-\frac{1}{a^2}\right)
$$

