1
$\begingroup$

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

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.