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when $x_1$ and $x_2$ are dependent, we know that $E[x_1x_2]^2 \neq E[x^2_1]E[x^2_2]$.

Is it possible to express $E[x_1x_2]^2$ in terms of $E[x^2_1]$ and $E[x^2_2]$?

I only know that $\quad var[x_i]=E[x^2_i]-E[x_i]^2$ and $cov[x_1,x_2]=E[x_1x_2]-E[x_1]E[x_2]$, but it doesn't help.

Lee
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  • The notation $E[x_1x_2]^{2}$ has been interpreted differently by different people in the answers. I believe the square is taken before the expectation. @Lee I think you should clarify this point. – Kavi Rama Murthy Jun 11 '18 at 10:11
  • @KaviRamaMurthy: I guess that the comparison of $E[x_1x_2]^2 $ and $ E[x^2_1]E[x^2_2]$ lifts any ambiguity. –  Jun 11 '18 at 11:55
  • If you see that first implication in the answer by Tony S. F. there is a clear mis-understanding of the notation. I agree that the notation by the OP is reasonable but it is better to add extra parenthesis. – Kavi Rama Murthy Jun 11 '18 at 12:03
  • @KaviRamaMurthy: Here by $E[x_1x_2]^2$ I mean $E[x_1x_2] \times E[x_1x_2]$ – Lee Jun 12 '18 at 02:35
  • In that case the answer by drhab is wrong. – Kavi Rama Murthy Jun 12 '18 at 05:29

3 Answers3

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Just square the relationship for covariance,

$$cov[x_1,x_2]+E[x_1]E[x_2] = E[x_1x_2]\\ \implies (cov[x_1,x_2]+E[x_1]E[x_2])^2 = E[x_1x_2]^2\\ \implies cov[x_1,x_2]^2 + 2cov[x_1,x_2]E[x_1]E[x_2] + E[x_1]^2E[x_2]^2 = E[x_1x_2]^2$$

Jürgen Sukumaran
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There is an expression for $\mathbb E[x_1x_2]^2$ in which $\mathbb Ex_1^2$ and $\mathbb Ex_2^2$ are present. See the answer of Tony for instance, and also you could "use":$$\mathbb E[x_1x_2]^2=\mathbb E[x_1x_2]^2-\mathbb Ex_1^2-\mathbb Ex_2^2+\mathbb Ex_1^2+\mathbb Ex_2^2$$

But purely an expression in terms of $\mathbb Ex_1^2$ and $\mathbb Ex_2^2$ and nothing else? No!

Suppose that there would indeed be some expression:$$\mathbb E[x_1x_2]^2=f(\mathbb Ex_1^2,\mathbb Ex_2^2)$$

Then e.g. if $x_1,x_2$ are iid then we would find:$$\mathbb Ex_1^4=\mathbb E[x_1x_1]^2=f(\mathbb Ex_1^2,\mathbb Ex_1^2)=f(\mathbb Ex_1^2,\mathbb Ex_2^2)=\mathbb E[x_1x_2]^2=\mathbb Ex_1^2\mathbb Ex_2^2=[\mathbb Ex_1^2]^2$$

or equivalently $\mathsf{Var}(x_1^2)=0$ so that $x_1^2$ is degenerated.

So the existence of such expressions allows us to prove that for all random variables $x_1$ that have finite $4$-th moment the random variable $x_1^2$ is degenerated (which is absurd).

drhab
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Without more information, it is not possible.

The dependency is precisely what makes $E[x_1x_2]^2$ different from $E[x^2_1]E[x^2_2]$, and without more information, all you can say is $$-\sqrt{E[x^2_1]E[x^2_2]}< E[x_1x_2] <\sqrt{E[x^2_1]E[x^2_2]}.$$

Knowing the covariance or the corrrelation coefficient, you could retrieve the difference between the two expressions.