What is the distribution of $P_M(M_B(t))$

$$M_X(t)= P_M(M_B(t))$$

$$P_M(s)= (1-q+qs)^2$$

$$M_B(t)= \frac{\beta}{\beta-t}$$

Where P(x) is the probability generating function and M(x) is a moment generating function

I identified M as $$M \sim binomial (2,q)$$ and $$B \sim exp (\beta)$$

but I don't know what is the distribution of X?

The solution tell $$X \sim BinomComp (2,q,F_B)$$ but I didn't understand what this distribution is.

I know what a mixture distribution is , I know conditional distributions but I've never seen something like this. Any explanation is very appreciated. Thank you!

• I got the answer $X = B_1+ B_2 + ... + B_N$ You can try to find M_X(t) and it will be $P_M(M_B(t))$ – Youssef Feb 10 at 0:13

You are asking what distribution corresponds to the $$M_X(t)$$. This is difficult to get unmotivated but in fact it is known that the composition of a probability generating function of an $$\mathbb{N}_0$$ random variable and an MGF (or CF) is the MGF of a random sum of independent random variables. To see this we see that if,

$$X = \sum_{i=1}^M B_i$$

Where $$B_i$$ are iid random variables distributed the same way $$B$$ is, and $$M$$ is an independent $$\mathbb{N}_0$$ random variable (which you identified as being $$\text{Binomial}(2,q)$$). Thus, $$S$$ is a random sum, to find the MGF of $$S$$ we see that,

$$M_X(t) = \mathbb{E} [e^{tX}] = \sum_{n \in \mathbb{N}_0} \mathbb{E}[e^{tX} \ | \ M = n] \cdot \mathbb{P}[M = n]$$

Where the second equality comes from the law of total expectation. Note that $$X$$ conditional on $$M$$ is just a sum of the $$B$$ distributed random variables

$$X \ | \ M = n \sim \sum_{i=1}^n B_i$$

Therefore we have,

$$\mathbb{E}[e^{tX} \ | \ M = n] = \mathbb{E}\left[e^{t\sum_{i=0}^n B_i} \right] = \mathbb{E} \left[\prod_{i=0}^n e^{tB_i} \right] = (\mathbb{E} e^{tB})^n$$

Note there is a subtle issue here when $$n = 0$$. This corresponds to a degenerate case of the random sum where $$X = 0$$ and so the MGF is identically $$1$$. Putting these facts together we obtain,

$$M_X(t) =\sum_{n \in \mathbb{N}_0} \mathbb{E}[e^{tX} \ | \ M = n] \cdot \mathbb{P}[M = n] = \sum_{n \in \mathbb{N}_0} (\mathbb{E} e^{tB})^n \cdot \mathbb{P}[M = n] = \mathbb{E}[(\mathbb{E} e^{tB})^M] = P_M(M_B(t))$$

Therefore we've proven that the distribution of $$X$$ is simply as described,

$$X = \sum_{i=1}^M B_i$$

To find this distribution explicitly we find the CDF of $$X$$ through the law of conditional probability where for $$x \geq 0$$,

$$\mathbb{P}[X \leq x] = \sum_{n=0}^2 \mathbb{P}[X \leq x \ | \ M = n] \cdot \mathbb{P}[M = n]$$

Expanding this out we have,

$$\mathbb{P}[X \leq x] = (1-q)^2+ 2q(1-q) \mathbb{P}[B_1 \leq x] + q^2 \mathbb{P}[B_1 + B_2 \leq x]$$

$$B_1 \sim \text{Exp}(\beta)$$ and since $$B_1, B_2$$ are independent exponential, their sum is (shape-rate parameterisation) $$B_1 + B_2 \sim \text{Gamma}(2, \beta)$$ distributed, and so we have after differentiating,

$$f_X(x) = 2q(1-q) \cdot \beta e^{-\beta x} + q^2 \cdot \beta^2 x e^{-\beta x}$$

Finally,

$$f_X(x) = \beta e^{-\beta x} (2q(1-q) + \beta q^2 x)$$