Jan 29, 2019 · From the definition of expected value and the probability mass function for the binomial distribution of n trials of probability of success p, we can demonstrate that our intuition matches with the fruits of mathematical rigor.

If X ~ B(n, p) and Y | X ~ B(X, q) (the conditional distribution of Y, given X), then Y is a simple binomial random variable with distribution Y ~ B(n, pq). For example, imagine throwing n balls to a basket U X and taking the balls that hit and throwing them to another basket U Y .

Dec 5, 2024 · Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$. Then the expectation of $X$ is given by: $\expect X = n p$ Proof 1. From the definition of expectation: $\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$ Thus:

Aug 11, 2023 · Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$. Then the variance of $X$ is given by: $\var X = n p \paren {1 - p}$ Proof 1. From the definition of Variance as Expectation of Square minus Square of Expectation: $\var X = \expect {X^2} - \paren {\expect X}^2$

Nov 1, 2012 · It is easy to verify that E() = p, so E(X) = np. You wrote down another expression for the mean. So the above argument shows that the combinatorial identity of your problem is correct. You can think of it as a mean proof of a combinatorial identity.

If a random variable X has a binomial distribution, we write X ~ B(n, p) (~ means ‘has distribution…’). n and p are known as the parameters of the distribution (n can be any integer greater than 0 and p can be any number between 0 and 1).

The mean and variance for the approximately normal distribution of X are np and np(1-p), identical to the mean and variance of the binomial(n,p) distribution. Similarly, the mean and variance for the approximately normal distribution of the sample proportion are p and (p(1-p)/n) .

Jun 5, 2020 · Definition of Binomial Distribution, with $p+q=1$: $$ \mathrm{E}(X)=\sum_{k=0}^{n} k\cdot P(X=k)=\sum_{k=0}^{n} k\left(\begin{array}{l}n \\ k\end{array}\right) p^{k} q^{n-k} $$ since for $k=0, k\left(\begin{array}{l}n \\ k\end{array}\right) p^{k} q^{n-k}=0$, we get $$ \sum_{k=1}^{n} k\left(\begin{array}{l}n \\ k\end{array}\right) p^{k} q^{n-k ...

Oct 19, 2019 · Closed 5 years ago. E(X) = np if X is a binomial random variable is the statement I have to prove. The definition of a binomial random variable tells us that P(X = h) = (n k)ph(1 − p)n − h. E(X) =∑a∈R aP(X.

(n-x)! x! (shorten to only x things)(remove duplicates) A combination takes the number of ways to make an ordered list of n elements (n!), shortens the list to exactly x elements ( by dividing this number by (n-x)!