0
$\begingroup$

Could someone let me know if this makes sense?

Let $X_1,...,X_n$ be independent Bernoulli random variables B(1,p) and let $S_n=X_1+...+X_n$.

Determine the conditional distribution of $X_1$ given that $S_n=r.$
Determine the conditional distribution of $S_k$, given that $S_n=r$ if $k<n.$


My attempt:
$X_1$ is a Bernoulli random variable so it can take the value $0$ with probability $1-p$ or $1$ with probability $p$. $S_n$ is a binomial random variable with parameters $n$ and $p$.

$P_{x_1=0|S_n=r}=\frac{P(S_n=r|x_1=0)P(x_1=0)}{P(S_n=r)}=\frac{{n-1 \choose r}p^r(1-p)^{n-r}(1-p)}{{n \choose r}p^r(1-p)^{n-r}}=\frac{(1-p)(n-r)}{n}$.

$P_{x_1=1|S_n=r}=\frac{P(S_n=r|x_1=1)P(x_1=1)}{P(S_n=r)}=\frac{{n-1 \choose r-1}p^r(1-p)^{n-r}p}{{n \choose r}p^r(1-p)^{n-r}}=\frac{pr}{n}$.
If this is correct, then I think I am able to determine the other conditional distribution ( I think it would simplify to a hypergeometric distribution).

$\endgroup$
  • $\begingroup$ Are the Bernoulli r.v.'s identically distributed? You never said so, but your proof assumes it. $\endgroup$ – Clement C. Apr 26 '18 at 2:15
  • $\begingroup$ @ClementC. Hmmmm well I guess I just assumed that. The prompt is exactly what I wrote above so ... If they're not identically distributed I cannot make the claim that $S_n$ is a binomial random variable with parameters n and p... $\endgroup$ – funmath Apr 26 '18 at 2:20
  • $\begingroup$ Indeed, in that case it is a Poisson Binomial Distribution. $\endgroup$ – Clement C. Apr 26 '18 at 2:21
  • $\begingroup$ Although I would assume they meant to imply the $X_i$'s are i.i.d. (?). You are sure it was written "independent" and not "i.i.d."? $\endgroup$ – Clement C. Apr 26 '18 at 2:22
  • $\begingroup$ @ClementC. Just checked! It says "....be independent Bernoulli random variables B(1,p)". So yes they have the same parameter p. $\endgroup$ – funmath Apr 26 '18 at 2:28
2
$\begingroup$

Almost. When given a value of $x_1$ you know what one from the $n$ results will be, and just need to consider how the remaining $n-1$ results will give $S_n-x_1$ successes.

If the random variables are independent and identically Bernoulli distributed with parameter $p$.

$$\begin{split}\mathsf P({x_1=0\mid S_n=r})&=\frac{\mathsf P(S_n-x_1=r-0\mid x_1=0)\cdot \mathsf P(x_1=0)}{\mathsf P(S_n=r)}\\&=\frac{{n-1 \choose r}p^r(1-p)^{\color{red}{n-1}-r}\cdot(1-p)}{{n \choose r}p^r(1-p)^{n-r}}\\&=\frac{(n-r)}{n}\end{split}$$

$$\begin{split}\mathsf P({x_1=1\mid S_n=r})&=\frac{\mathsf P(S_n-x_1=r-1\mid x_1=1)\cdot\mathsf P(x_1=1)}{\mathsf P(S_n=r)}\\&=\frac{{n-1 \choose \color{red}{r-1}}p^{\color{red}{r-1}}(1-p)^{n-r}\cdot p}{{n \choose r}p^r(1-p)^{n-r}}\\&=\frac{r}{n}\end{split}$$

Reality check: $\frac{n-r}n+\frac r n=1$

Intuition check: If the $r$ successes had been spread among the $n$ samples without bias, then the probability that the first sample was among the successes will be $r/n$.

Now similarly: $$\mathsf P(S_k=t\mid S_n=r) = \dfrac{\mathsf P(S_n-S_k=r-t\mid S_k=t)\cdot \mathsf P(S_k=t)}{\mathsf P(S_n=r)}$$

Use similar intuition to anticipate your result.

$\endgroup$
  • $\begingroup$ Thank you! I got it! $\endgroup$ – funmath Apr 26 '18 at 3:48

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.