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Consider IID $X_1, \ldots, X_n$. Consider $Y = X_1 + \ldots + X_n$

What is the formal reason why $Y \neq nX_i$?

I tried to explain this to a younger student by saying something like "they're random variables, so you can't add them in that way because the randomness means they aren't guaranteed to have the same realization. If they were guaranteed to be IID AND also have the same value, then you can add them."

Then I give an example where $X \in \{0, 1\}$. If we use $Y = X_1 + X_2$, then $ \in \{0,1,2 \}$, but if we use $Y = 2X_1$, then $Y \in \{0, 2\}$.

student010101
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  • Random variables are measurable functions by definition. They are equal only when they are equal as functions, i,e $X(\omega)=Y(\omega)$ for all $\omega\in\Omega$. This is not the same thing as saying they have the same distribution. So $X_1+...+X_n$ is not the same thing as $X_1+...+X_1$. – Mark Sep 20 '20 at 18:31
  • Here $X_1, ... X_n$ are functions and $+$ doesn't have the usual meaning of pointwise summation, indeed $Y=X_1+X_2$ in this context means $Y(\omega_1, \omega_2)=X_1(\omega_1) + X_2(\omega_2)$. – Giovanni Barbarani Sep 20 '20 at 18:36
  • @GiovanniBarbarani Usually random variables are assumed to be defined on the same sample space and then it is usual pointwise sum of functions. However, these are still distinct functions in the sum. – Mark Sep 20 '20 at 18:40
  • What is $N$ in this context? Is it different from $n$, the number of random variables? – Rivers McForge Sep 20 '20 at 18:45
  • @Mark from a measure theory point of view here $Y$ is a function of the random vector $(X_1, X_2)$ defined on the product space $\Omega^2$. – Giovanni Barbarani Sep 20 '20 at 18:48
  • @RiversMcForge No, it is the same. I changed all $N$ to $n$ in latest edit. – student010101 Sep 20 '20 at 18:57
  • @GiovanniBarbarani I always define $(X,Y)$ as a random variable $\Omega\to\mathbb{R^2}$ by $(X,Y)(\omega)=(X(\omega), Y(\omega))$. – Mark Sep 20 '20 at 19:46

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Let's take your example where the two rv's take value in $\{0;1\}$...(with some probability :))

If you do $\{0;1\} \times 2=\{0;2\}$ you suppose that, when for example the first rv=1 also the second rv is 1. But it is not true....both rv's can take the values zero-one indipendently...with a defined probability (defined by their probability law)

Thus you can have the following cases

  1. $(X,Y)=(0;0)$

  2. $(X,Y)=(0;1)$

  3. $(X,Y)=(1;0)$

  4. $(X,Y)=(1;1)$

tommik
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