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)

$nZ$ looks a strange thing to write. – Henry Jan 04 '23 at 00:00 — Henry, Jan 04 '23 at 00:00

score 3 · Answer 1 · answered Jan 03 '23 at 21:28

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.

answered Jan 03 '23 at 21:28

Christophe Leuridan

8,223

1

The range of a random variable needs not to be $\mathbb R$. – Sam Wong Jan 03 '23 at 21:32
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@Sam Wong It should be at least a vector space. Otherwise, that is the meaning of $(X_1+\cdots+X_n)/n$? – Christophe Leuridan Jan 03 '23 at 21:43
Didn't notice that the title is about $X_1+...+X_n$ but the actual question is about $(X_1+...+X_n)/n$. In the latter case we do need a vector space, though the "minimal" underlying field can be $\mathbb Q$. – Sam Wong Jan 04 '23 at 10:11
It's not enough to just be a vector space of course; you need some sense of measurability which needs to be compatible with the vector space structure; for instance you could consider topological vector spaces equipped with the Borel sigma algeba. – Aditya Dhawan Jan 04 '23 at 10:31
@AdityaDhawan Sure. Random variables are just measurable functions, which has already implied the underlying space of the range should have some sort of measurable structure (compatible with the scalar multiplication $1/n$). Otherwise, it makes no sense to talk about measurable functions. – Sam Wong Jan 04 '23 at 12:42

score 2 · Answer 2 · edited Jan 03 '23 at 21:40

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 $X_n$.

edited Jan 03 '23 at 21:40

Christophe Leuridan

8,223

answered Jan 03 '23 at 21:09

DanishChef

419

So in both cases(Xi's are all discrete and Xi's are all continuous), they are still random variables, right? (By induction ) 2. Is my method correct?(I mean the method start at let X1+X2+...+Xn=nZ is also a random variable)

Elizabeth Amanda

Jan 03 '23 at 21:21

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@ChristopheLeuridan It is given in the question that the $X_i$ are independent. – Théophile Jan 03 '23 at 21:30

Sam Wong · Accepted Answer · 2023-01-04T10:19:15.227

0

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.

edited Jan 04 '23 at 10:19

answered Jan 03 '23 at 21:30

Sam Wong

2,277

According to your answer, in both cases (X1+X2+...+Xn)/n is random variable, right? (Two cases are X1,X2,...Xn are all i.i.d and discrete, and X1,X2,...Xn are all i.i.d and continuous.) – Elizabeth Amanda Jan 04 '23 at 11:26
@ElizabethAmanda Yes. My argument doesn't need the i.i.d condition and apply to ALL random variables, i.e. we don't care if these random variables are continuous or discrete. – Sam Wong Jan 04 '23 at 11:33
I got it. Thank you very much. – Elizabeth Amanda Jan 04 '23 at 12:34

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

3 Answers3