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I’m completing the first part of my university course about Probability & Statistics.

Sometimes, I’ve noticed that hypotheses and theses in our proofs are kinda “hidden”, and as a consequence I’m struggling to understand certain points.

I’ll write the proposition stated in the title here below.

Proposition to prove

Let $X, Y$ be two random variables defined on the same probability space $(\Omega, \mathcal{F}, P)$ and which are admitting finite variance. Thus, $X + Y$ admits finite variance and $\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y) + 2 \cdot \mathbb{E}[(X - \mathbb{E}(X)) \cdot (Y - \mathbb{E}(Y))]$.

What I understand

I’ve understood the reasoning behind the first part of the proposition involving the fact that $X + Y$ admits finite variance, but I can’t comprehend the last part, as specified below.

What I struggle to understand

During the lecture, we proved the second part as follows.

Proof of the second part:

$$\text{Var}(X + Y) := \mathbb{E}[{(X + Y - \mathbb{E}(X + Y))}^2]$$

Since both $\mathbb{E}(X)$ and $\mathbb{E}(Y)$ are finite because $X, Y$ admit finite variance as hypothesis, then $\mathbb{E}(X + Y) = \mathbb{E}(X) + \mathbb{E}(Y)$.

$$\mathbb{E}[{(X + Y - \mathbb{E}(X + Y))}^2] = \mathbb{E}[{(X -\mathbb{E}(X) + Y - \mathbb{E}(Y))}^2]$$

We can now expand the square.

$$\mathbb{E}[{(X -\mathbb{E}(X) + Y - \mathbb{E}(Y))}^2] = \\ = \mathbb{E}[{(X -\mathbb{E}(X))}^2 + {(Y -\mathbb{E}(Y))}^2 + 2 \cdot (X -\mathbb{E}(X)) \cdot (Y -\mathbb{E}(Y))]$$

Since all the terms admit finite expected value, we can break the internal sum in a sum of expected values, thus the thesis.

QED

The crucial point is in the last assumption.

In fact, consider the (expanded) product $(X -\mathbb{E}(X)) \cdot (Y -\mathbb{E}(Y))$. It is clearly true that the terms

  • $X \cdot \mathbb{E}(Y)$
  • $Y \cdot \mathbb{E}(X)$
  • $\mathbb{E}(X) \cdot \mathbb{E}(Y)$

admit finite expected value because of the hypotheses, but, generally speaking, $\mathbb{E}(X \cdot Y) \neq \mathbb{E}(X) \cdot \mathbb{E}(Y)$. The equality does hold if $X, Y$ are independent.

Our lecturer suggested that this last part of the proposition can easily be proved using the information that $X, Y$ admit finite variance, but I can’t seem to find a proper way to show that.

Basically, the only detail that I don’t know to solve is that $\mathbb{E}(X \cdot Y)$ is finite. If I can prove that, then I can apply the linearity of the expected value as reported in the proof and conclude the thesis.

Additional information

Notice that I’m not studying a Pure Mathematics or Physics curricula as reported in the tags, so I’m not looking for a pretty formal proof, but anyway I would be glad to understand everything properly in order to be prepared for my exam.

Thanks for your time.

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2 Answers 2

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Just a hint: you can try to use (in conjunction with the monotony of the expectation) the following inequality: $$|x||y|\leq \cfrac{1}{2}(x^2+y^2)$$ which holds for any real number $x$ and $y$.

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  • $\begingroup$ That’s a good advice. But what if $X \cdot Y$ wasn’t a random variable? Can I be sure that their product is a discrete or continuous random variable? $\endgroup$ Commented Apr 9, 2021 at 13:00
  • $\begingroup$ The product of two (real-valued) random variables is always a random variable (but giving rigorous sens to this statement needs a bit of mathematical formalism). And you don't need to know if $$X\cdot Y$$ is discrete or continuous to apply the inequality to get your result. Well, in fact it depends on the definition of the expectation you've been given (and the properties you know about it). Without knowing exactly the content of your course, it is difficult to be more specific. $\endgroup$
    – Cal
    Commented Apr 9, 2021 at 13:10
  • $\begingroup$ I don’t know if this can help you, but we introduced the concept of expected value in a not-too-precise mathematical way. What I’m trying to say is that even though the course is held by the Maths Departement, it is not an extremely formal course. Other information are: we didn’t introduced Borel measures (which I know are kinda connected), neither proved this general fact about $X \cdot Y$ which you stated. What I’m aware of is that I am allowed to apply (most of) the properties of $\mathbb{E}(\cdot)$ when the random variables involved admit finite expected value. (linearity, monotony, ...) $\endgroup$ Commented Apr 9, 2021 at 13:28
  • $\begingroup$ Well, so, you just need to 'accept' that $X\cdot Y$ is a random variable (until you have a complete course on measure/probability theory in which you will be given a precise meaning of what a "random variable" is...). In any case, you can substitute $x$ and $y$ in the inequality above by $X$ and $Y$ because there exists no possible value that $X$ and $Y$ can take such that this inequality won't be verified (well, it's not a fully rigorous argument we have here: we need to be able to give a precise meaning to $Z_1\leq Z_2$ for two random variables $Z_1$ and $Z_2$...), then use the monotony. $\endgroup$
    – Cal
    Commented Apr 9, 2021 at 13:50
  • $\begingroup$ I think your explanation is the best I can get for now. I don’t know if my degree will provide a Measure Theory course, or at least it won’t be a problem to worry about until then. Apart from that, I’m not quite sure if Measure Theory is involved in AI & Machine Learning, because that’s the specific degree I’ll try to pursue in the next years. $\endgroup$ Commented Apr 10, 2021 at 7:52
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May I suggest you a different but (probabilty) a way to clear up your dubts?

$$\mathbb{V}[X+Y]=\mathbb{E}[X+Y]^2-\mathbb{E}^2[X+Y]=$$

$$=\mathbb{E}[X^2]+\mathbb{E}[Y^2]+2\mathbb{E}[XY]-\{\mathbb{E}[X]+\mathbb{E}[Y]\}^2=$$

$$=\underbrace{\mathbb{E}[X^2]-\mathbb{E}^2[X]}_{=\mathbb{V}[X]}+\underbrace{\mathbb{E}[Y^2]-\mathbb{E}^2[Y]}_{=\mathbb{V}[Y]}+2\{\underbrace{\mathbb{E}[XY]-\mathbb{E}[X]\cdot\mathbb{E}[Y]}_{\text{covariance}}\}=$$

$$=\mathbb{V}[X]+\mathbb{V}[Y]+2 \mathbb{Cov} [XY]$$

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  • $\begingroup$ I think we are returning to the starting point: you supposed that $\mathbb{E}(XY)$ is finite in order to break the sum of expected values in the very first equality, but that is what I’m trying to prove. $\endgroup$ Commented Apr 9, 2021 at 12:57
  • $\begingroup$ @fabio-ciani-polimi: suppose that $E(XY)=\infty$. Then also $V(X+Y)=\infty$ but this is a contraddiction as you stated that the two rv's are in $L^2$ $\endgroup$
    – tommik
    Commented Apr 9, 2021 at 13:14
  • $\begingroup$ Well, if $\mathbb{E}(XY) = +\infty$ I wouldn’t be allowed to break the terms into a sum of expected values, or am I wrong? $\endgroup$ Commented Apr 9, 2021 at 13:24

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