As been said in this question iid variables, do they need to have the same mean and variance? the answer is yes, but I'm looking for the proof for that for general random variables, not for continously or discrete random variables.
So the question is, why do iid random variables have the same parameters, mean, variance?
Thanks in advance Chaim
"iid" stands for "independent, identically distributed". If random variables are identically distributed, they must come from the same distribution. A distribution that has a mean and variance has a unique mean and variance. If random variables come from a common distribution, they are identically distributed, and they must have a common mean and variance.
To be more mathematical, let's say $X_n$ is a sequence of random variables, independent and identically distributed from a distribution I'll call $\mathcal{D}$: $$ X_n\sim\mathcal{D} $$
The distribution $\mathcal{D}$ has an associated density function, let's call it $f_D(x)$, with mean given by $$ \mu = \int_{-\infty}^{\infty}xf_D(x)dx $$ and variance given by $$ \sigma^2 = \int_{-\infty}^{\infty}(\mu-x)^2f_D(x)dx $$
Note that with the appropriate use of delta functions in $f_D$, this can represent a probability mass function for discrete variables as well, so this applies in either case.
By definition $$ \mathbb{E}[X_n]=\int_{-\infty}^{\infty}xf_D(x)dx=\mu $$ and $$ \text{Var}[X_n]=\mathbb{E}[(X_n-\mu)^2] = \int_{-\infty}^{\infty}(\mu-x)^2f_D(x)dx=\sigma^2 $$ So, if variables are iid, they have the same mean and variance. This extends to all higher order moments as well. The key word is "identical".