# Proof of Expected Value Property for product of Independent Variables

I keep seeing this property come up for two random variables $$X,Y$$ in a probability space $$(\Omega,\mathcal{M}, P)$$. If two random variables are independent, then $$\mathbb{E}[XY] = \mathbb{E}[X] \mathbb{E}[Y]$$. But I have confusion on the space and measure that the expected value is calculated on.

If we start with the definition that two random variables are independent if for any $$A,B\in\mathcal{M}$$ we have that the events $$\{X\in A\},\{Y\in B\}$$ are independent. From this definition we can compute that if $$X,Y$$ are simple functions with values $$x_1,\ldots x_n,y_1,\ldots y_m$$ then $$\mathbb{E}[XY] =\sum_{i,j}x_iy_j P(X = x_i, Y = y_j) =\Big(\sum_{i}x_i P(X=x_i)\Big)\Big(\sum_{j}x_i P(Y=y_j)\Big)=\mathbb{E}[X] \mathbb{E}[Y]$$

I would assume from here that we would probably need $$X,Y\in L^2(P)$$ to allow us to interchange the order of limits as needed above? We would then want to perform a normal limit argument by choosing any sequence of simple random variables $$X_\alpha, Y_\beta$$ such that $$X_\alpha\to X$$ and $$Y_\beta \to Y$$ to show that since $$\mathbb{E}[X_\alpha Y_\beta] = \mathbb{E}[X_\alpha] \mathbb{E}[Y_\beta]$$ then $$\mathbb{E}[XY] = \mathbb{E}[X] \mathbb{E}[Y]$$.

The above roughly makes sense to me until I start seeing definitions like this https://en.wikipedia.org/wiki/Independence_(probability_theory)#Two_random_variables
where it seems that the definition is based on the joint probability from the probability spaces. So if $$Q = P\otimes P$$ then the expected values above should be written as $$\mathbb{E}_Q[XY] = \mathbb{E}_P[X] \mathbb{E}_P[Y]$$

So is the first notation $$\mathbb{E}[XY] = \mathbb{E}[X] \mathbb{E}[Y]$$ just an abuse of notation for what is occurring? So talking about independent random variables will infer that we are workingin the product space?

• You definitely need some form of Fubini-Tonelli theorem to conclude that $E[XY]=E[X]E[Y]$. In general, the result holds if $X,Y\geq 0$ or if $E(|X|),E(|Y|)<\infty$. Jun 8, 2020 at 17:16

Assume that $$X,Y \in \mathcal L^1$$.

Let us denote the distribution of $$X$$ by $$\mu$$ and the distribution of $$Y$$ by $$\nu$$. Further denote the joint distribution of $$X,Y$$ by $$\pi$$. The probability measures $$\mu$$ and $$\nu$$ are therefore measures on $$\Bbb R$$, where $$\pi$$ is a measure on the product space $$\Bbb R \times \Bbb R$$ with $$\sigma$$-algebra $$\mathcal B( \Bbb R ) \otimes \mathcal B( \Bbb R )$$. The point is, that the quantities $$\Bbb E [X] , \Bbb E [Y] , \Bbb E [XY]$$, if existing, only depend on $$\mu, \nu , \pi$$, respectively, since by integrating with the Pushforward measures $$\mu ,\nu , \pi$$ we have

$$\Bbb E [X]=\int_\Omega X(\omega) d P(\omega ) = \int_{\Bbb R} x d \mu (x), \quad \Bbb E [Y]=\int_\Omega Y(\omega) d P(\omega ) = \int_{\Bbb R} y d \nu (y)$$ and $$\Bbb E [XY] = \int_\Omega X(\omega)Y(\omega) d P(\omega ) = \int_{\Bbb R^2} xy d \pi (x,y)$$

Now note that $$X,Y$$ are independent if and only if $$\pi = \mu \otimes \nu$$ by the definition and uniqueness of the product measure on $$\sigma$$-finite measure spaces.

But this means that, if $$X,Y$$ are independent, by the Fubini-Tonelli theorem (the german page treats the general case with more arbitrary measures) we have

$$\int_{\Bbb R^2} xy d \pi (x,y) = \int_{\Bbb R} \int_{\Bbb R} xy d\mu (x) d \nu (y) = \int_{\Bbb R} x d \mu (x) \int_{\Bbb R} y d \nu (y)$$ since the righthand side above exists, due to $$X,Y \in \mathcal L^1$$.

• So we do not need apriori to know that $XY$ is integrable since Fubini-Tonelli shows us its existence since $X,Y\in\mathcal{L}^1$ Jun 9, 2020 at 23:28
• If they are independent, yes. Jun 10, 2020 at 9:35
• does this even require that $X$ and $Y$ be defined on the same probability space? Dec 29, 2021 at 23:08

This answer is based on the book Measures, Integrals and Martingales, by René L. Schilling. The author derives the Expected Value Property before defining product measures and product $$\sigma$$-algebras. So indeed, you can derive this property without the use of Fubini's theorem. Here is how Schilling does it:

First, let's start with the notion of independence. Let $$(\Omega, \mathcal A, P)$$ be a probability space and $$\mathcal B, \mathcal C \subset \mathcal A$$ be two sub-$$\sigma$$-algebras. We say that $$\mathcal B$$ and $$\mathcal C$$ are independent if $$P(B \cap C)=P(B)P(C) \, \forall B \in \mathcal B, C \in \mathcal C$$

With that in mind, we move on to the Expected Value Property. So, let's first assume that $$u = \mathbb I_B$$ and $$w = \mathbb I_C$$. Because of the independence we have

$$\int uw dP = P(A \cap B) = P(A)P(B) = \int u dP \int w dP$$

Now, for positive simple functions, such that $$u = \sum_j \alpha_j \mathbb I_{B_j}$$ and $$w = \sum_i \beta_i \mathbb I_{C_i}$$ we get

$$\int uw dP = \sum_{j,i}\alpha_j \beta_i \int \mathbb I_{B_j} \mathbb I_{C_i} dP =$$

$$= \sum_{j,i}\alpha_j P(B_j \cap C_i) = \sum_{j,i}\alpha_j P(B_j)P(C_i) =$$

$$= \left( \sum_j P(B_j) \right) \left( \sum_i P(C_i) \right) = \int u dP \int w dP$$

Now, for $$u \in \mathcal M^+(\mathcal B)$$ and $$w \in \mathcal M^+(\mathcal C)$$, we can use approximating simple functions $$u_n \uparrow u$$ and $$w_n \uparrow w$$, then, using the Monotone Convergence Theorem: $$\int uw dP = lim_{n\to\infty} \int u_n w_n dP = lim_{n\to\infty} \int u_n dP \int w_n dP = \int u dP \int w dP$$

Finally, if $$u \in L^1(\mathcal B)$$ and $$w \in L^1(\mathcal C)$$, then $$u \cdot w$$ is integrable, because $$\int \mid u w \mid dP \leq \int \mid u \mid dP \int \mid w \mid dP < + \infty$$ So, we split the positive and negative parts of each $$u$$ and $$w$$, and finally

$$\int uw dP = \int u dP \int w dP$$