Mathematics, 29.07.2019 09:30, bella51032
Arandom variable x that assumes the values x1, x2, . . , xk is called a discrete uniform random variable if its probability mass function is f(x) = 1 k for all of x1, x2, . . , xk and 0 otherwise. find the mean and variance of x.
Answers: 1
following a discrete distribution, the expectation of ![\mathbb E[X]=\displaystyle\sum_x x\,f_X(x)](/tpl/images/0146/1701/1ece3.png)
is the PMF of 
![\mathbb E[X]=\displaystyle\sum_x x\,f_X(x)=\sum_{i=1}^k x_i\,f_X(x_i)=\frac1k\sum_{i=1}^k x_i](/tpl/images/0146/1701/c1ab5.png)
![\mathbb V[X]=\mathbb E[(X-\mathbb E[X])^2]=\mathbb E[X^2]-\mathbb E[X]^2](/tpl/images/0146/1701/03689.png)
![(\mathbb E[X^2])](/tpl/images/0146/1701/0a510.png)
we run into the same issue as before - we can't find a "complete" result - but we do get![\mathbb E[X^2]=\displaystyle\sum_x x^2\,f_X(x)=\frac1k\sum_{i=1}^k{x_i}^2](/tpl/images/0146/1701/80b78.png)
![\mathbb V[X]=\displaystyle\frac1k\sum_{i=1}^k{x_i}^2-\left(\frac1k\sum_{i=1}^kx_i\right)^2](/tpl/images/0146/1701/7cd78.png)
![\mathbb E[X]^2](/tpl/images/0146/1701/05f5d.png)
a bit. Denoting the set 
, we can write
![\mathbb V[X]=\displaystyle\left(\frac1k-\frac1{k^2}\right)\sum_{i=1}^k{x_i}^2-\frac1{k^2}\sum_{i,j\in K,\,i\neq j}x_ix_j](/tpl/images/0146/1701/a83f7.png)
![\mathbb V[X]=\displaystyle\frac{k-1}{k^2}\sum_{i=1}^k{x_i}^2-\frac1{k^2}\sum_{i,j\in K,\,i\neq j}x_ix_j](/tpl/images/0146/1701/55ef8.png)
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Arandom variable x that assumes the values x1, x2, . . , xk is called a discrete uniform random var...
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