# Compound binomial distribution distributed as binomial

Suppose we have independent family of random variables $$\{Y\}_{i\in\mathbb{N}}\cup\{N\}$$, where $$Y$$s are identically distributed.
Next consider a sum of random number of random variables $$W_N\equiv\sum_{i=1}^NY_i$$, where $$N\sim Binomial(n,q)$$ ($$n>1$$ is a number or trials, and $$q$$ is success probability). Distribution of $$Y$$s is given by: $$\mathbb{P}(Y_1=1)=\alpha,\quad\mathbb{P}(Y_1=2)=1-\alpha\quad(\alpha\in(0,1)).$$ Additionaly I have information that $$q=0.36$$ and I am asked about a value of $$\alpha$$, for which $$W_N$$ is distributed as binomial. I have a hint to use concept of Moment Generating Function (further represented as $$M_{r.v.}(\cdot)$$). So I made deriviation: $$M_{W_N}(t)=\mathbb{E}(e^{tW_N})=\\\mathbb{E}(\mathbb{E}(e^{tW_N}|N))=\\\mathbb{E}(\sum_{i=1}^\infty\mathbb{E}(e^{tW_N}|N=n)\mathbf{1}_{\{n\}}(N))=\\\mathbb{E}(\sum_{i=1}^\infty\mathbb{E}(e^{tW_n}|N=n)\mathbf{1}_{\{n\}}(N))=\\\sum_{i=1}^\infty M_{W_n}(t)\mathbb{P}(N=n)=\\\sum_{i=1}^\infty [M_{Y_1}(t)]^n\mathbb{P}(N=n)=\\\mathbb{E}([M_{Y_1}(t)]^N)=\\\mathbb{E}(e^{N\ln(M_{Y_1}(t))})=M_N(\ln(M_{Y_1}(t))).$$ Next I calculated $$M_{Y_1}(\cdot)$$ and plugged in $$M_N(t)=(1-q-qe^t)^n$$: $$M_{Y_1}(t) = e^t\alpha+e^{2t}(1-\alpha)\\M_{W_N}(t)=(1-q-q(e^t\alpha+e^{2t}(1-\alpha)))^n.$$ To obtain Moment Generating Function of binomial random variable, it must be in a form $$(1-Q+Qe^t)^m$$ for some $$Q\in(0,1), m\in\mathbb{N}$$. Here I have trouble to go further: by the fact that $$Y$$s take values in $$\{1,2\}$$, I am concluding that $$N\leq W_N\leq 2N$$ and $$m=2n$$. Even if my reasoning is correct, I would prefer some formal mathematical translation of this argument (which I cannot grasp).
But then when I am proceeding and equating polynomials, I am getting $$\alpha>1$$; steps below: $$(1-q-q(e^t\alpha+e^{2t}(1-\alpha)))^n=(1-Q-Qe^t)^{2n}\quad (x\equiv e^t)\\(1-q-q(x\alpha+x^2(1-\alpha)))=(1-Q)^2-2(1-Q)Qx+Q^2x^2\\(1-q)-q\alpha x-q(1-\alpha)x^2=(1-Q)^2-2(1-Q)Qx+Q^2x^2\rightarrow\\1-q=(1-Q)^2\land q\alpha=2(1-Q)Q\land-q(1-\alpha)=Q^2.$$ Solving, I obtain equation: $$q^2\alpha^2-4q(1-q)\alpha+4(1-q)q=0.$$ Solving for $$\alpha$$ gives solutions of the form: $$\alpha^*=\frac{2[(1-q)\pm\sqrt{(1-q)(1-2q)}]}{q}.$$ After substituting $$q=0.36$$ I obtain 2 numbers bigger than $$1$$, which is incorrect, since $$\alpha$$ is probability of event $$\{Y_1=1\}$$.
I would be grateful for any hints!

• Your calculations should produce $\alpha =1$ as a solution as you would then have $W_N=N$ which is normally distributed Feb 18 at 0:52

We can recognize that the sampling process has resemblance to a multinomial distribution and we take the sum of two categories. Alternatively we can see it as a sum of a categorical distribution with probabilities

$$P(X_i = x_i) = \begin{cases} 1-q & \quad \text{if \quad x_i=0}\\ q\alpha & \quad \text{if \quad x_i=1}\\ q(1-\alpha)& \quad \text{if \quad x_i=2} \end{cases}$$

And your variable is $$W_n = \sum_{i=1}^n X_i$$

Now the trick is to

1. figure out the moment generating function of $$X_i$$
2. apply the rules for moment generating functions of a sum of independent variables
3. compare it with the formula for the moment generating function of a binomial distribution

Without applying moment generating functions, we can already see that the case $$\alpha = 1$$ reduces $$X_i$$ to a Bernoulli variable for which we know that the sum is a binomial distribution. My intuition tells me that it is the only case, but with the moment generating functions you can prove it and I guess that the exercise also wants you to practice with such functions.

• I didn't read your question thoroughly. It looks like you already correctly derived the moment generating function and the step is to realize that the form $(1-Q+Qe^t)^m$ is only obtained with $(1-q-q(e^t\alpha+e^{2t}(1-\alpha)))^n$ if $\alpha = 1$ in which case the $e^{2t}$ term disappears. Feb 18 at 14:54