If $X_1,X_2,…,X_n$ are i.i.d random variables and are all discrete/continuous, then is $(X_1+X_2+…X_n)/n$ also a random variable?
My attempt: For continuous type, I guess it is a random variable. Since $Z=X+Y$ is a random variable, we can view $X_1+X_2+…+X_n=nZ$, then we can get the CDF of it. After we got the CDF, we can get the PDF. But I stuck on the CDF Step, because at here it is in higher dimensions, so we can not use the classical method in two dimensions to get the pdf.
And I am also not sure about the discrete case. Could someone explain more to me?
Please give me the answer about two cases. ($X_1,X_2,…,X_n$ are all discrete and $X_1,X_2,…,X_n$ are all continuous)
Moreover, if we just apply the definition of random variables(transfer the event to a real number), then maybe in both cases. They are random variables. But here we are trying to transfer multiple events? Will the sample space(collection of all outcomes) change to higher dimensions?(like the case in joint pmf/pdf)
Best Answer
By definition, the random variables with values on $\mathbb{R}$ defined one a same probability space $(\Omega,\mathcal{A},P)$ are the measurable functions from $(\Omega,\mathcal{A})$ to $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. The collection of random variables is stable under linear combinations, and products.