$X_1, X_2, .....X_n$ is a colletion of Random Variables. They are said to be $\textit{multiplicative system}$ if, for any $1 \leq k \leq n$ and for any set of $k$ indices $1 \leq i_1 < i_2 ....i_k \leq n$ -

$$E[X_{i_1}X_{i_2}.....X_{i_k}] = 0 $$ I need to prove that if $X_1, X_2, ....X_n$ are multiplicative system then, $$E\bigg[ \prod_{i=1}^n(a_iX_i + b_i) \bigg] = \prod_{i=1}^{n}b_i$$ over any choice of real constants $a_1, a_2, .....a_n$ and $b_1, b_2, ....b_n$.


1 Answer 1


Let us consider first an example with only $n=3$ random variables. So, we will have $X_{1},X_{2},X_{3}$ are a multiplicative system.

And we will prove that

$$\mathbb{E}[\prod_{i=1}^{3}(a_{i}X_{i}+b_{i})] =\prod_{i=1}^{3}b_{i}$$

We have the product $(a_{1}X_{1}+b_{1})*(a_{2}X_{2}+b_{2})*(a_{3}X_{3}+b_{3})$, where if we expand it we will have that it is equal to

$$a_{1}a_{2}a_{3}X_{1}X_{2}X_{3}+b_{2}a_{1}a_{3}X_{1}X_{3}+b_{3}a_{1}a_{2}X_{1}X_{2}+b_{2}b_{3}a_{1}X_{1}+\\b_{1}a_{2}a_{3}X_{2}X_{3}+b_{1}b_{3}a_{2}X_{2}+b_{1}b_{2}a_{3}X_{3}+b_{1}b_{2}b_{3} \ \ \ (**)$$

We results in a summation of $8$ products. However, only one of these products doesn't include a subset of the $X_{1},X_{2},X_{3}$ and this is the $b_{1},b_{2},b_{3}$.

Hence, from the definition of the Multiplicative System we will have that $\mathbb{E}[X_{1}X_{2}X_{3}]=\mathbb{E}[X_{1}X_{3}]=\mathbb{E}[X_{1}X_{2}]=\mathbb{E}[X_{1}]=\mathbb{E}[X_{2}X_{3}]=\mathbb{E}[X_{2}]=\mathbb{E}[X_{3}]=0$

So, by taking the expected value over $(**)$ we have that the only term that has non-zero expected value is $\mathbb{E}[b_{1}b_{2}b_{3}]=\prod_{i=1}^{3}b_{i}$.

Now, let consider the case were we have $n=4$ random variables, again if we expand the product $\prod_{i=1}^{4}(a_{i}X_{i}+b_{i})$ it will result into a summation of products, where only one product will not include a subset of $X_{1},X_{2},X_{3},X_{4}$ and that will be the $b_{1}b_{2}b_{3}b_{4}$. In that kind of analogy in the case where we have $X_{1},X_{2},...,X_{n}$ random variables, in the expantion of the product $\prod_{i=1}^{n}(a_{i}X_{i}+b_{i})$ there will be only one term that it is free of any subset of $X_{1},X_{2},...,X_{n}$ and that would be the $\prod_{i=1}^{n}b_{i}$.

  • $\begingroup$ how to extend this result for $n$ Random Variables? $\endgroup$ Commented May 9, 2021 at 10:26
  • $\begingroup$ I'll edit that on the anwser $\endgroup$
    – Fiodor1234
    Commented May 9, 2021 at 10:28
  • $\begingroup$ @SeñoraPenn The key point is that for any choice of $n$ in the product only one term will be free of $X_{i}$ $\endgroup$
    – Fiodor1234
    Commented May 9, 2021 at 10:35
  • $\begingroup$ From the multiplicative property we have that all subsets of $X_{i}$ are equal to zero, i.e. $\mathbb{E}[X_{1}]=0,\mathbb{E}[X_{2}]=0,\mathbb{E}[X_{3}]=0$ and $\mathbb{E}[X_{1}X_{2}]=0,\mathbb{E}[X_{1}X_{3}]=0,\mathbb{E}[X_{2}X_{3}]=0$ and $\mathbb{E}[X_{1}X_{2}X_{3}]=0$ and because all of the are equal to zero I just said that they should also be equal. It is not necessary to say that they are equal you can only state that each one of them is simply equal to zero. $\endgroup$
    – Fiodor1234
    Commented May 9, 2021 at 10:41

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