If $X$ has a binomial distribution with parameters $n$ and $p$, show that  \begin{align*} \textbf{E}\left(\frac{1}{1+X}\right) = \frac{1-(1-p)^{n+1}}{(n+1)p} \end{align*}  MY ATTEMPT  Since $X\sim\text{Binomial}(n,p)$, we have \begin{align*} \textbf{E}\left(\frac{1}{1+X}\right) &amp; = \sum_{x=0}^{n}{n\choose x}\frac{p^{x}(1-p)^{n-x}}{1+x} = (1-p)^{n}\sum_{x=0}^{n}{n\choose x}\frac{1}{1+x}\left(\frac{p}{1-p}\right)^{x}\\\\ &amp; = \frac{(1-p)^{n+1}}{p}\sum_{x=0}^{n}{n\choose x}\frac{1}{1+x}\left(\frac{p}{1-p}\right)^{x+1} \end{align*}  This is as far as I can get. Could someone help me out? Thanks in advance!
45975
The problem:  X is a binomial random variable, find $E[\frac{1}{X+1}]$ n and p are not given  PDF for a binomial distribution is $\binom{n}{k}p^k(1-p)^{n-k}$  Expected value is $\sum{x_ip(x_i)}$  But this is where I get stuck, I'm really rusty on my statistics and I'm not sure exactly how to structure it in the next step? I think I want to get the form of the following out of the summation  $\sum _{k=0}^{n} \binom{n}{k}p^k(1-p)^{n-k} = (p + 1 - p)^n = 1$  But I'm not sure if it should look like  $\sum \frac{1}{xp(x)+1} $  and if it should where to go from here?
27949
$$\sum_{n=0}^\infty \binom{n}{k}p^k(1-p)^{n-k}\frac{\lambda^ne^{-\lambda}}{n!} = \frac{(\lambda p)^ke^{-\lambda p}}{k!}$$  That is, given a random variable $X$ with Poisson distribution $X \sim \operatorname{Poisson}(\lambda)$, and, given $X = n$, random variable $Y$ is distributed by a binomial such that $Y \sim B(n,p): p \in [0,1]$. Given $n$, $P(Y = k|X = n) = \binom{n}{k}p^k(1-p)^{n-k}$. Given $\lambda$, $P(X=n) = \dfrac{\lambda^ne^{-\lambda}}{n!}$. Therefore, $P(Y = k|\lambda) = \sum_{n=0}^\infty P(Y = k | X = n)P(X = n | \lambda) = \sum_{n=0}^\infty \binom{n}{k}p^k(1-p)^{n-k}\dfrac{\lambda^ne^{-\lambda}}{n!}$. What properties and identities can be used to demonstrate the equality above?
91289
Suppose $X$ is binomial $B(n,p)$. How can I find the probability that $X$ is even ?  I know $$P(X = k ) = \frac{ n!}{(n-k)!k!} p^k(1-p)^{n-k} $$  where $X=1,....,n$. Are they just asking to find $P(X = 2m )$ for some $m &gt; 0$ ?
11589
I have this problem on a textbook that doesn't have a solution. It is:  Let $$f(x)=\frac{\binom{r}{x} \binom{N-r}{n-x}}{\binom{N}{n}}\;,$$ and keep $p=\dfrac{r}{N}$ fixed. Prove that $$\lim_{N \rightarrow \infty} f(x)=\binom{n}{x} p^x (1-p)^{n-x}\;.$$  Although I can find lots of examples using the binomial to approximate the hypergeometric for very large values of $N$, I couldn't find a full proof of this online.
26296
Using the definition of the binomial distribution, I obtain that: $$\Psi (t) = (pe^t+q)^n $$ I then compute $\Psi ' = npe^t(pe^t+q)^{n-1}$  I then evaluated this at $\Psi'(0)$ and got $\Psi'(0)=np(p+q)^{n-1}$  and so $E[X]$ which I know is $np$.   How do I then differentiate $\Psi'(t)$ again, the $n-1$ is throwing me off?
54957
Sum of binomial product $\displaystyle \sum^{n}_{r=0}\binom{p}{r}\binom{q}{r}\binom{n+r}{p+q}$   Simplifying $\displaystyle \frac{p!}{r!\cdot (p-r)!} \cdot \frac{q!}{r!\cdot (q-r)!}\cdot \frac{(n+r)!}{(p+q)! \cdot (n+r-p-q)!}$.  Could some help me with this, thanks
22029
$$P(X_1 &lt; X_2 &lt; X_3)= \frac{(1-p_1)(1-p_2)p_2p_3^2}{(1-p_2p_3)(1-p_1p_2p_3)}$$   To be frank I do not know where to start with this question, I would like an idea to get me going, or better yet an answer. This was on an earlier exam, I would like to be ready if something like this were to come up again tomorrow. What makes this a little bit more difficult than usual is the fact that geometric distribution is  a discrete variable..
34581
I came across this question as an exercise, had a brief idea, but didn't know how to proceed.     Let $X \sim N(0, 1)$. What is the p.d.f of $|X|$ ?   I know the final p.d.f looks just like the right half of the original pdf, but extended vertically for a factor of 2.  Could someone please show mathematically what the p.d.f of variable |X| is going to be, and explain briefly? Thank you.
94432
How would I find the sum below?    $$ \sum_{k=0}^{n} {\frac{1}{k+1} {n \choose k}}=?$$  I need help starting on the problem. Would I first need to utilize the Binomial Theorem proof to get started on it?    Thank you.
76179
Let X and Y have a trinomial distribution with join probability function:  $ f(x,y) = \frac{n!}{x!y!(n-x-y)!}p^x q^y (1-p-q)^{n-x-y}; x=0,1,...n, y=0,1,...n$  Let $T=X+Y$  I can't seem to figure out where to go exactly. Here's where I am at:  $\sum_{x=0}^{n} \frac{n!}{x!y!(n-x-y)!}p^x q^y (1-p-q)^{n-x-y}\mbox{ substitute in Y=T-X}$ $\sum_{x=0}^{n} \frac{n!}{x!(t-x)!(n-t)!}p^x q^{t-x} (1-p-q)^{n-t}$   I think I could I figure out the algebra.. if I could just see what to do with the summation. I know I have to involve a t somewhere, but I'm not sure. Maybe somebody could just show me how to set up my summation correctly with the substitution?
41990
Let $X_1,X_2,...$ be a sequence of i.i.d Bernoulli random variables   with parameter $p$. Let $N$ be a Poisson random variable with with   parameter $\lambda$ which is independet of the $X_i.$      (a) Find the probability generating function of $Z=\sum_{i=1}^{N}X_i.$      (b) Use (a) to identify the distribution of $Z$.   The pgf of a sum of i.i.d random variables is the product of their respective pgf's so:  $$G_Z(s)=G_{X_1}(s)\cdot G_{X_2}(s)\cdot...\cdot G_{X_N}(s),$$  and sice the $X_{i}$ are Bernoulli distributed with parameter $p$ we have that  $$G_Z(s)=(1-p+ps)^N.$$  I'm not sure how to proceed from here. I can't express my $G_Z(s)$ in terms of a random variable, so what should I do with the $N$?
76097
I am trying to define the probability distribution of $Z$ such as $Z = X_1\cdot X_2$ where $X_1$ and $X_2$ are two independent and identically exponentially distributed variables.  $$P(X_1=x) = \lambda e^{-\lambda x}$$ $$P(X_2=x) = \lambda e^{-\lambda x}$$  I tried something like that….  $$P(Z=x) = P(X_1 = a) \cdot P(X_2 = x/a)$$  Is this first equation correct?  $$P(Z=x) = \lambda e^{-\lambda a} \cdot \lambda e^{-\lambda x/a} = \lambda^2 e^{-\lambda (a+x/a)}$$  I doesn't feel like I found the solution as the dummy variable $a$ remains in the final result. Can you please help me solving this problem?
75580
I was hoping to show that  $$\left(1-\frac{x}{n}+o\left(\frac{2x}{n}\right)\right)^n \xrightarrow{n\to\infty} e^{-x}$$  which would be just fine without the little-$o$.  Trying binomial formula:  $$\left(1-\frac{x}{n}+o\left(\frac{2x}{n}\right)\right)^n = \sum_{k=0}^n \binom{n}{k} \left(1-\frac{x}{n}\right)^{n-k} \left( o\left( \frac{2x}{n} \right)  \right)^k$$  Yet it doesn't look like it leads me somewhere where I could use  $$\frac{o\left( \frac{2x}{n} \right)}{\frac{2x}{n}}\xrightarrow{n\to\infty} 0$$  What is the proper way to handle it?  Edited after the comment by Claude Leibovici.
80337
Consider the Bernoulli polynomials $B_n(x)$ given by the expansion  $$\frac{te^{xt}}{e^t-1} = \sum\limits_{n=0}^{\infty}B_n(x)\frac{t^n}{n!}.$$  I want to prove the identity $$B_n(1-x)=(-1)^nB_n(x).$$  This formula is given in a lot of books, but unfortunately without any proof. Therefore I'm wondering how to prove it. Do you have any idea?  I tried for instance:  $\sum\limits_{n=0}^{\infty}B_n(1-x)\frac{t^n}{n!}=\frac{te^{-xt}}{e^t-1}e^t=\left(\sum\limits_{n=0}^{\infty}B_n(-x)\frac{t^n}{n!}\right)\sum\limits_{n=0}^{\infty}\frac{t^n}{n!}=\sum\limits_{n=0}^{\infty}\left(\sum\limits_{l=0}^n\binom{n}{l}B_{l}(-x)  \right)\frac{t^n}{n!}$.  Hence  $B_n(1-x)=\sum\limits_{l=0}^n\binom{n}{l}B_{l}(-x)$.  But I do not know how to proceed.  On the other hand I know that $B_n(x+1)-B_n(x)=n x^{n-1}$. Hence  $B_{n}(1-x)=B_{n}(-x)-n(-x)^{n-1}$. But again, I do not know how I can go further.   Hopefully someone can help me.  Best regards
114867
How to get a closed form for $$\sum_{k=0}^{n-p}\binom{n}{k}\binom{n}{p+k}\;?$$  I tried to write binominal in term of gamma function but I got no result. What is your suggestion to solve the problem?
59439
Is there any way of determining if $\binom{n}{k} \equiv 0\pmod{n}$.  Note that I am aware of the case when $n =p$ a prime.  Other than that there does not seem to be any sort of pattern (I checked up to $n=50$).  Are there any known special cases where the problem becomes easier?  As a place to start I was thinking of using $e_p(n!)$ defined as:  $$e_p(n!) = \sum_{k=1}^{\infty}\left \lfloor\frac{n}{p^k}\right \rfloor$$  Which counts the exponent of $p$ in $n!$ (Legendre's theorem I believe?)  Then knowing the prime factorization of $n$ perhaps we can determine if these primes appear more times in the numerator of $\binom{n}{k}$ than the denominator.  Essentially I am looking to see if this method has any traction to it and what other types of research have been done on this problem (along with any proofs of results) before.  Thanks!
25151
I need to express $$1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb  + \frac{1}{n + 1}\binom{n}{n}$$ in a simplified form.  So I used the identity $$(1+x)^n=1 + \binom{n}{1}x + \binom{n}{2}x^2 + \dotsb   + \binom{n}{n}x^n$$ Now on integrating both sides and putting $x=1$.  I am getting $$\frac{2^{n+1}}{n+1}$$ is equal to the given expression.But the answer in my book is  $$\frac{2^{n+1}-1}{n+1}.$$ Where does that -1 term in the numerator come from?