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I need a derivation of mean and variance formula for multinomial distribution. I tried to prove the formula, but I don't know what is meaning of expected value and variance in multinomial distribution. The formula for variance and mean is given as below in wikipedia: $ E({X}_{i})=n{p}_{i}\phantom{\rule{0ex}{0ex}} \;\; Variance({X}_{i})=n{p}_{i}(1-{p}_{i}) $ What do these equations indicate in definition of expected value?(in $E(X)=\sum _{x}^{}x\cdot p(x)$)

How can these equations be proved?

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they are the expectation and variance of the Outcome $i$ of the distribution. The single outcome is distributed as a Binomial $\text{Bin}(n;p_i)$ thus mean and variance are well known (and easy to prove)

Mean and variance of the multinomial are expressed by a vector and a matrix, respectively...in wikipedia link all is well explained IMHO

to prove these indicators simply observe that a binomial $S_n\sim \text{Bin}(n;p_i)$ is the sum of $n$ iid bernulli thus

$$\mathbb{E}(S_n)=\mathbb{E}\left(\sum_i X_i \right)=n\mathbb{E}(X_i)=np_i$$

$$\mathbb{V}(S_n)=\mathbb{V}\left(\sum_i X_i \right)=n\mathbb{V}(X_i)=np_i(1-p_i)$$


$$E(X)=\sum_{x=0}^{n}x\binom{n}{x}p^xq^{n-x}=\sum_{x=0}^{n}x\frac{n!}{x!(n-x)!}p^xq^{n-x}=$$

$$=np\sum_{x=1}^{n}\frac{(n-1)!}{(x-1)!(n-x)!}p^{x-1}q^{n-x}=$$

now set $y=x-1$and $m=n-1$ and you get

$$=np\underbrace{\sum_{y=0}^{m}\binom{m}{y}p^yq^{m-y}}_{=1}=np$$

to calculate the variance first similarly calculate $E(X^2)$ setting

$$x^2=x+x(x-1)$$

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  • $\begingroup$ can you prove the mean and variance formula by using $E(X)=\sum _{x}^{}x\cdot p(x)$ $\endgroup$
    – joshua
    Jun 24, 2021 at 8:28
  • $\begingroup$ @joshua: added a proof...it is also possible to get an analytical one. Did you really retract you upvote? :( $\endgroup$
    – tommik
    Jun 24, 2021 at 8:38
  • $\begingroup$ but I need a proof which contains definition of expected value. I wrote the formula the comment before. You have used the linearity of expectation. But my question differs your answer $\endgroup$
    – joshua
    Jun 24, 2021 at 8:49
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    $\begingroup$ @joshua: linearity of expectation is a good way to proceed. I used your formula for the bernulli and then used the linearity of expectation. Anyway now I added another proof....hope this helps $\endgroup$
    – tommik
    Jun 24, 2021 at 8:55
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    $\begingroup$ @joshua : probabily it is my fault but I am not able to explain you the problem. The formula you used is the one of a single outocome $X_i$ that is binomial distributed. You cannot apply your univariate formula to get mean and variance of a multivariate distribution as the multinomial is.... $\endgroup$
    – tommik
    Jun 24, 2021 at 9:15

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