On the normal mixture $X\sim\mathcal N(NY,NY(1-Y))$, $Y\sim\operatorname{Beta}(\alpha,\beta)$

Question

Suppose I generate a random variable by first selecting from $Y \sim \operatorname{Beta}(\alpha, \beta)$

I then use the result to generate a random variable distributed as $X \sim \operatorname{Normal}(NY,NY(1-Y))$ for some fixed $N$.

Clearly $X$ is a mixture of normals, but is $X$ itself a normal distribution, or any other well known distribution?

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Below is an attempt using MGF as per comment suggestion.

We have: $M_x = E_Y[M_{x|y}]$

Now $$M_{x|y}(t) = \exp(Ny t + \frac{1}{2}Ny(1-y) t^2) = \exp(Nyt(1 + \frac{(1-y)t}{2}))$$

Now presumably $$E_Y[M_{x|y}] = \int_{0}^{1} M_{x|y} f(y)dy = \int_{0}^{1} exp(Nyt(1 + \frac{(1-y)t}{2})) \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)} y^{\alpha-1}(1-y)^{\beta-1}dy$$

Now the moment generating function of the beta distribution is given by,

$$M_b(t,\alpha,\beta) = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta))}\int_{0}^{1}e^{ty}y^{\alpha-1}(1-y)^{\beta-1}dy$$

Thus,

$$M_b(Nt(1 + \frac{t}{2}),\alpha,\beta) = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta))}\int_{0}^{1}e^{Nt(1 + \frac{t}{2})y}y^{\alpha-1}(1-y)^{\beta-1}dy$$

Answer 1

Clearly $X$ is a mixture of normals, but is $X$ itself a normal distribution, or any other well known distribution?

As you have already pointed out $X$ cannot be normal as it is a normal mixture; however, conditioning $X$ on $Y$ does give a normally distributed random variable since $$ X|Y\sim\mathcal N(nY,nY(1-y)). $$

The random variable $X$ is an example of a normal location-scale mixture. The Wikipedia article in the link gives some examples of well-known examples of this type of mixture model; however, none of the given examples match your specific model.

As for the moments of $X$ note that since $X|Y\sim\mathcal N(nY,nY(1-y))$ we can write the moments of $X$ as $$ \mathsf EX^k=\mathsf E(\mathsf E(X^k|Y)). $$ Using equation (18) in this arXiv article allows us to write this expression as $$ \mathsf EX^k=\mathsf E\left(\left(\frac{i\sqrt{nY(1-Y)}}{\sqrt 2}\right)^kH_k\left(-i\frac{nY}{\sqrt{2nY(1-Y)}}\right)\right), $$ where the expectation is taken w.r.t. $Y\sim\operatorname{Beta}(\alpha,\beta)$. Using the explicit expression for the Hermite polynomial this becomes $$ \begin{aligned} \mathsf EX^k &=\mathsf E\left(\left(\frac{i\sqrt{nY(1-Y)}}{\sqrt 2}\right)^kk!\sum_{\ell=0}^{\lfloor k/2\rfloor}\frac{(-1)^\ell}{\ell!(k-2\ell)!}2^{k-2\ell}\left(-i\frac{nY}{\sqrt{2nY(1-Y)}}\right)^{k-2\ell}\right)\\ &=k!\sum_{\ell=0}^{\lfloor k/2\rfloor}\frac{2^{-\ell}n^{k-\ell}}{\ell!(k-2\ell)!} \mathsf E\left(Y^{k-\ell}(1-Y)^\ell\right)\\ &=k!\sum_{\ell=0}^{\lfloor k/2\rfloor}\frac{2^{-\ell}n^{k-\ell}}{\ell!(k-2\ell)!} \mathsf E\left(Y^{k-\ell}(1-Y)^\ell\right). \end{aligned} $$ Knowing the moments of the beta distribution we obtain the final result $$ \mathsf EX^k =k!\sum_{\ell=0}^{\lfloor k/2\rfloor}\frac{2^{-\ell}n^{k-\ell}}{\ell!(k-2\ell)!} \frac{\operatorname B(\alpha+k-\ell,\beta+\ell)}{\operatorname B(\alpha,\beta)}, $$ which may (possibly) be written in terms of hypergeometric functions.

The analytical expression for the moments can be implemented in Mathematica as

EX[k_] := 
  k! Sum[(2^-l n^(k - l))/(l! (k - 2 l)!)
      Beta[\[Alpha] + k - l, \[Beta] + l]/Beta[\[Alpha], \[Beta]], {l,
      0, Floor[k/2]}];
Table[{k, FullSimplify[FunctionExpand[EX[k]]]}, {k, 0, 3}];
TeXForm[%]

Here is a table of the first several

$$ \left( \begin{array}{cc} k & \mathsf EX^k \\ 0 & 1 \\ 1 & \frac{\alpha n}{\alpha +\beta } \\ 2 & \frac{\alpha n (\beta +\alpha n+n)}{(\alpha +\beta ) (\alpha +\beta +1)} \\ 3 & \frac{\alpha (\alpha +1) n^2 (3 \beta +(\alpha +2) n)}{(\alpha +\beta ) (\alpha +\beta +1) (\alpha +\beta +2)} \\ \end{array} \right) $$