1
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1answer
41 views

How do you calculate a binomial distribution with k > R as opposed to k = R

I'm given the formula: $\displaystyle P(X = k; n, p) = \binom {n}{k} * p^k * q^{n-k}$ And we need to work out the binomial coefficient by hand, instead of using C(n,r). So I have a question: "Some ...
0
votes
1answer
25 views

Likelihood of Two Binomial Distributed RV's

We are given that Let X1~Bin(n1 = 34, p1) and X2~Bin(n2 = 56, p2) In general, what is the likelihood, L(p1, p2) = f (X1, X2 | p1, p2) for the data X1 and X2 I believe that I am supposed to use a ...
0
votes
1answer
38 views

Binomial distribution convergence

Let $Y \sim\binom{n}{\pi}$ Suppose $n \rightarrow \infty$ and $\pi \rightarrow 0$ such that $n\pi \rightarrow \mu$, where $\mu$ is a constant. derive the limiting distribution of Y. $f_Y(y)= ...
1
vote
2answers
148 views

Can I get a little help proving equality between a summation and integral?

Prove$$\sum_{k=0}^x \binom{n}{k}p^{k}(1-p)^{n-k} =(n-x)\binom{n}{x}\int_{0}^{1-p}t^{n-x-1}(1-t)^{x}dt.$$ Can someone show me the steps please? Here is the hint my book gave me: "Integrate by parts ...
0
votes
2answers
295 views

Multivariate Hypergeometric Distribution With Wildcard

In the wikipedia link http://en.wikipedia.org/wiki/Hypergeometric_distribution it is obvious you can do things like "I draw 5 cards from a deck of 50 cards where there are 10 cards that equate to ...
1
vote
2answers
327 views

Probability of getting >70% in exam with 50 yes/no questions

In a paper containing 50 yes/no questions, I am trying to find the probability of getting 70%. Using binomial distribution, $$P(X\ge70\%)=\sum_{k=25}^{50} \binom{50}{k}\left(\frac{1}{2}\right)^{50}$$ ...
0
votes
1answer
99 views

Are binomial distribution answers expressed as percentages?

I have a question on an answer to a binomial distribution question. The chances of getting $100\%$ in a test is $8.881784197 \cdot 10^{-16}$. Is that actually a percentage? Making it ...
0
votes
1answer
49 views

Sample size for Wilsons confidence interval

Consider a story ranking website in which the ranking is crowd sourced from the number of up votes and down votes received by a story. The score is computed as the lower bound Wilson's algorithm. ...
2
votes
1answer
225 views

Going from binomial distribution to Poisson distribution

Why does the Poisson distribution $$\!f(k; \lambda)= \Pr(X=k)= \frac{\lambda^k \exp{(-\lambda})}{k!}$$ contain the exponential function $\exp$, while its relation to the binomial distribution would ...
3
votes
1answer
88 views

is the approximation of the sum true?

Someone commented under my question Calculation of the moments using Hypergeometric distribution that $$ \sum_{k=0}^l\frac{{l \choose k}{2n-l \choose n-k}(2k-l)^q}{{2n\choose n}}\sim \sum_{k=0}^l ...
1
vote
0answers
63 views

calculation of the sum using idea of one answer

I am wondering if the sum (the $q$-th moment) in my question Calculation of the moments using Hypergeometric distribution can be calculated using idea in Evaluating 'combinatorial' sum ? ...
5
votes
1answer
493 views

Calculation of the moments using Hypergeometric distribution

Let vector $a\in 2n $ is such that first $l$ of its coordinates are $1$ and the rest are $0$ ($a=(1,\ldots, 1,0, \ldots, 0)$). Let $\pi$ be $k$-th permutation of set $\{1, \ldots, 2n\}$. Define ...
2
votes
1answer
310 views

Evaluating 'combinatorial' sum

Help me please to calculate the following sum. I have seen such kind of formulas in the papers related to combinatorics, specifically 'trees'. I am curious how to calculate or approximate this sum: ...
4
votes
1answer
130 views

Calculate $\sum_{i=1}^{[\frac{\sqrt n}{2}]}{n\choose i}$

It is known that $\sum_{i=1}^n {n \choose i}=2^n$. I am wondering what would be the sum if we change the upper limit to $\sqrt n/2$, i. e. How to calculate$$\sum_{i=1}^{[\frac{\sqrt n}{2}]}{n \choose ...
4
votes
2answers
374 views

computation of the sum

I am having trouble to compute the following sum: $$ \sum_{k=0}^n(n-2k)^p \frac{{n \choose k}{2m-n \choose m-k}}{{2m \choose m}} $$ Here $p\geq 2$. To simplify the question, we can even assume that ...
2
votes
5answers
280 views

summation of x * (y choose x) binomial coefficients

What does this summation simplify to? $$ \sum_{x=0}^{y} \frac{x}{x!(y-x)!} $$ I was able to realize that it is equivalent to the summation of $x\dbinom{y}{x}$ if you divide and multiply by $y!$, but ...