Is $(X_1+X_2+...+X_n)/n$ a random variable?

Question

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)

Answer 1

Short answer: Yes.

I don’t know what is your definition of random variables. A random variable is just a measurable function from a underlying space called measure space which has a probability measure to another measurable space. If you add up a finite number of measurable functions together, the sum is still a measurable function, where the domain and codomain are still the same. Therefore, the sum is a random variable.

Edit: Well, the title is about $X_1+...+X_n$ but the actual question is about $(X_1+...+X_n)/n$. In the latter case, to make sense of the meaning of $1/n$, we do need the codomain be in a vector space with the underlying field being $\mathbb Q$ (or any field extension of $\mathbb Q$). Moreover, for the latter case, we need one more result that the finite product of measurable functions is measurable, cf. here. At last, we notice that $1/n$ is measurable as a constant function. The domain and comdomain remain the same. Therefore, $(X_1+...+X_n)/n$ is also a random variable.

Answer 2

It's the same in the discrete case, it's just induction.

If $X_1, X_2, \dots, X_n$ are random variables, then $X_1 + X_2$ is a sum of two random variables, thus a random variable. Then $(X_1 + X_2) + X_3$ is a random variable as well, and so is $(X_1 + X_2 + X_3) + X_4$, and so on. Then when you divide a random variable (like $X_1 + X_2 + \dots + X_n$) by $n$, it's still a random variable.

And since the $X_1,\ldots,X_n$ are assumed to be independent, the distribution of their sum is the convolution product of the individual distributions of $X_1$, ... , of $X_n$.

Answer 3

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.