Consider random variables $P$ and $X$ where $P \sim Uniform(0,1)$ and $X|P \sim Binomial (1, P)$. For any $s \in [0,1]$, calculate both $\mathbb{P}(P \leq s, X = 0)$ and $\mathbb{P}(P \leq s, X = 1)$.

I have an intuition on what to do, but I am having trouble justifying it rigorously and would like some assistance. This is what I have:

For $s \in [0,1]$,

$$\mathbb{P}(P \leq s, X=0) = \int_0^s \mathbb{P}(X=0|P=p)f_P(p)dp$$ $$=\int_0 ^s \frac{1}{1-0}(1-p)dp = s - \frac{s^2}{2}$$

The first part is from the PDF of a uniform distribution, the second part of the integral comes from drawing a "tail" for $X$. Yet since $X$ is discrete, I am not sure if I am allowed to move it inside the integral. I understand that there may be many flaws to my approach, so please enlighten me.

Calculate the conditional distribution of $P$ given $X=1$.

My attempt: $$\mathbb{P}(P \leq s | X = k) = \frac{\mathbb{P}(P \leq s, X = k)}{\mathbb{P}(X=k)}$$

From previous problem, we know that $\mathbb{P}(P \leq s, X = k) = \frac{s^2}{2}$ when $k=1$. Then $$\mathbb{P}(X = 1) = \int_0 ^1 \mathbb{P}(X=1 | P = p) f_P(p)dp=\int_0^1 p \frac{1}{1-0}dp=\frac{1}{2}$$

Thus $$\mathbb{P}(P\leq s|X=1) = \frac{s^2}{2} \cdot \frac{1}{\frac{1}{2}}=s^2$$ and finally, $$f_{P|X=1}(s)=\frac{d}{ds} \mathbb{P}(P \leq s |X=1)=\frac{d}{ds}s^2=2s$$


1 Answer 1


We should take the joint distribution $f_{X,P}(x,p)=\mathbb P(X=x|p)\cdot f_P(p)$ and integrate/sum it along the region defined by $X=0$ and $P<s$: $$\mathbb P(P<s, X=0)=\int_0^s \mathbb P(X=0|p)\cdot f_P(p)dp=\int_0^s(1-p)dp=s-\frac{s^2}{2}$$

Your final result is correct, but the step $\mathbb{P}(P \leq s, X=0) = \mathbb{P}_P (P\leq s) \cdot \mathbb{P}_{X | P=p}(X=0)$ is not correct. The actual result from the product rule is $\mathbb{P}(P \leq s, X=0) = \mathbb{P}_P (P\leq s) \cdot \mathbb{P}_{X | P\leq s}(X=0)$, which not really useful here. The integral you wrote after it is right though, since it is the integral of the joint probability distribution over the region of interest.

  • $\begingroup$ Thank you for your explanation. Going off of what you did, would you mind checking my work on the posterior distribution as well? $\endgroup$ Oct 13, 2020 at 4:08
  • 1
    $\begingroup$ @CharlieCornell It looks correct to me $\endgroup$
    – PedroSebe
    Oct 13, 2020 at 4:42

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