1
$\begingroup$

I have $X_i (i=1,...,n)$ which has $\text{Bernoulli}(p)$ or $\text{binomial}(1,p)$ distribution and each $X_i$ is independent with $X_i=1$ with probability $p$ and $X_i=0$ with probability $(1-p)$. I know that $S_n=X_1+\dots+X_n$ follows $\text{binomial}(n,p)$ distribution.

I am having trouble finding the conditional probability of $X_1=1$ given $S_n=k$. I am using the formula: $$P(X =x|Y =y)= P(X =x\text{ and }Y =y)/P(Y = y).$$ But I am having trouble finding the numerator. I am new to probability so any hints and suggestions would be appreciated.

$\endgroup$
  • $\begingroup$ Use MathJax.... $\endgroup$ – Aqua Aug 29 '17 at 22:28
  • $\begingroup$ This time I did it for you but you should typeset your question using LaTeX. $\endgroup$ – Arash Aug 30 '17 at 19:59
3
$\begingroup$

The probability that $S_n = k$ is of course $\binom{n}{k}p^k(1-p)^{n-k}$.

For $X_1$ to be $1$ and also $S_n = k$ two things must be true: $X_1 = 1$ (probability $p$) and $k-1$ among the remaining $n-1$ $X$-es must be $1$ (probability $\binom{n-1}{k-1}p^{k-1}(1-p)^{n-1-(k-1)}$).

So the conditional probability you want is $$ \frac{p \binom{n-1}{k-1}p^{k-1}(1-p)^{n-k}}{\binom{n}{k}p^k(1-p)^{n-k}}= \frac{\binom{n-1}{k-1}}{\binom{n}{k}}=\frac{(n-1)!}{(k-1)!(n-k)!}\frac{k!(n-k)!}{n!} = \frac{k}{n} $$

Now you can slap your forhead, because it was trivial to see that $k$ out of the $n$ variables were $1$, so the probability of any specific variable being $1$ would have to be $\frac{k}{n}$. All that work for such an easy result!

$\endgroup$
2
$\begingroup$

\begin{align*} k&= E [ S_n | S_n =k]\\ & = \sum_{j=1}^n E[ X_j | S_n =k]\\ & = n P(X_1=1|S_n =k)\\ & \Rightarrow P(X_1=1|S_n=k) = \frac kn\end{align*}

$\endgroup$
0
$\begingroup$

$\mathsf P(X_1{=}1\mid S_n{=}k)$ is the probability that a particular trial (the first) is a success when given that exactly $k$ among the $n$ trials are successes.

Symmetry of the situation should immediately suggests this probability is $k/n$.   It is.


We can also do $\newcommand{\hide}[1]{\bbox[0.5ex,lemonchiffon]{\phantom{#1}}} \mathsf P(X_1{=}1\mid \sum_{j=1}^n X_j{=}k)~{~=~\dfrac{\mathsf P(X_1{=}1)\,\mathsf P(\sum_{j=2}^n X_j{=}k{-}1)}{\mathsf P(\sum_{j=1}^n X_j{=}k)}\\~=~ \dfrac{{p}\cdot\hide{ \binom {n-1}{k-1} p^{k-1}(1-p)^{n-k}}}{\binom n k p^k(1-p)^{n-k}} \\~=~ \dfrac{k}{n} }$

$\endgroup$

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.