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So, I have the following conditions:

$\text{Let } X, Y \text{ be random variables, } |X| < C, \ |Y| < C, \\ \text{and } \mathbb{E}X^nY^m = \mathbb{E}X^n \ \cdot \mathbb{E}Y^m \quad \forall n, m \in \mathbb{N} \ \cup \{0\}$

I understand that this is a very popular question and has been asked here many times and there are many different answers to it, for example Bounded random variables $X,Y$ satisfying $\mathbb{E}(X^mY^n) = \mathbb{E}(X^m)\mathbb{E}(Y^n)$ for every $m, n\in\mathbb{N}$ are independent

I also don't want to use a statement like here: Criterion for independency of random variables

But I came up with a solution through characteristic functions, although I think that someone had already proposed it before me, but one point is not clear to me.

I want to consider characteristic function: $$\varphi_{X,Y}(s,t) = \mathbb{E}e^{i(sX+tY)} = \mathbb{E}e^{isX}e^{itY}$$

Also I want to say, that $$e^{isX} = \sum_{k=0}^\infty \frac{(isX)^k}{k!}, e^{itY} = \sum_{\ell=0}^\infty \frac{(itY)^\ell}{\ell!}$$

Next, I use the linearity of mathematical expectation: $$\mathbb{E}\left(\sum_{k=0}^n \frac{(isX)^k}{k!} \cdot \sum_{\ell=0}^m \frac{(itY)^\ell}{\ell!}\right) = \mathbb{E}\sum_{k=0}^n \frac{(isX)^k}{k!} \cdot \mathbb{E}\sum_{\ell=0}^\infty \frac{(itY)^\ell}{\ell!} $$

And then we need to make the limit transition, but I don't understand well, which theorem should I use? Generally speaking, I really want to apply Lebesgue's theorem, but I don't understand well what the majority function will be in this case, and I also don't understand well why there is a limitation of random variables? Of course, I would like to say that as such I can take: $|e^{itX}| \leqslant1 \ (= 1)$ And $1$ is integrable, but one thing confuses me is that I don't use the limitation of quantities anywhere.

But I'm not really sure about this inequality, and I don't really understand why it's integrable. Please tell me how to properly justify this subtle point with the ultimate transition?

Michael Hardy
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Wither_1422883
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  • What exactly is your question? Is it if $|e^{itX}|\leq 1$ holds? Then note that $|e^{itX}| = 1$, since $|...|$ is the complex absolute value. Or is your question something else? – Red Mar 12 '25 at 21:48
  • @Red I don't quite understand how to make the limit transition accurately (which theorem should I use?) and also why I need bounded values? – Wither_1422883 Mar 12 '25 at 21:55
  • I thought you said you already came up with a solution? If so then what is this exactly? In any event at least one of the links you posted uses characteristic functions--and posts additional links--so I have to close as a duplicate. – Mike Mar 12 '25 at 22:06
  • @Mike So the limit transition is not justified there and why it can be done, but in the link that you threw off, they just said that you can try to use the characteristic function. – Wither_1422883 Mar 12 '25 at 22:18
  • Let me try again as my last response was muddled [I apologize]. Here is where I am confused, and I cannot be the only one. You said in your OP that you "came up with a solution" but now you are saying that you need to make the limit transition which you do not understand well. Those two things seem to contradict each other. So you mean you came up with an approach and not a solution? And it would be good mentioning in your OP how this all ties to the previous questions posted here. – Mike Mar 12 '25 at 22:28
  • "Linearity" doesn't mean $\operatorname E(UV) = \operatorname E(U) \operatorname E(V)$ when $U,V$ are independent random variables. Linearity means that sums and scalar multiples are preserved. (But it is true that $\operatorname E(UV) = \operatorname E(U) \operatorname E(V)$ when $U,V$ are independent random variables, when both expectations exist.) – Michael Hardy Mar 13 '25 at 00:17
  • Use the Fubini's theorem – Mason Mar 13 '25 at 02:42

1 Answers1

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If I understood your question correctly, then we can make the following assessment: $$\left| \sum\limits^n_{k=0}\frac{(itX)^k}{k!} \right| \leqslant \sum\limits^n_{k=0} \left|\frac{(itX)^k}{k!}\right| = \sum\limits^n_{k=0}\frac{|(itX)^k|}{k!} = \sum\limits^n_{k=0}\frac{(|tX|)^k}{k!} \leqslant \sum\limits^n_{k=0}\frac{(|t|C)^k}{k!} \leqslant \sum\limits^{\infty}_{k=0}\frac{(|t|C)^k}{k!} = e^{|t|C}$$

Similarly, we get:$\left| \sum\limits^m_{l=0}\frac{(isY)^l}{l!} \right| \leqslant e^{|s|C}$ and $\left| \sum\limits^n_{k=0}\sum\limits^m_{l=0}\frac{(itX)^k}{k!}\frac{(isY)^l}{l!} \right| \leqslant e^{C(|t|+|s|)}$

Then for Lebesgue's theorem, as the majority functions, we take $g_1(t,s) = e^{C(|t|+|s|)}, \ g_2(t) = e^{C(|t|}, \ g_3(s) = e^{C(|s|}$

And $\mathbb{E}g_1, \ \mathbb{E}g_2, \ \mathbb{E}g_3 < \infty \quad \forall s,t \in \mathbb{R}$

And now you can make the limit transition.

Bogdan Witcher
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