# Covariance of a joint distribution: Factoring implies Independence discrepancy

Given the joint density: $$f(x,y)=\frac{16}{x^5y^6}, x,y\geq 1$$

We are asked to find the Cov$(X,Y)$. The solution manual says that Cov$(X,Y)=0$ because of factoring $f(x,y)=\frac{4}{x^5}\cdot \frac{4}{y^6}$.

However, using Cov$(X,Y)=<XY>-<X><Y>$, I checked the Expectations both by hand and through Mathematica:

$$<X>=\int _1^\infty \int _1^\infty x f(x,y) dx dy=\frac{16}{15}$$ $$<Y>=\int _1^\infty \int _1^\infty y f(x,y) dx dy=1$$ $$<XY>=\int _1^\infty \int _1^\infty x y f(x,y) dx dy=\frac{4}{3}$$

so that Cov$(X,Y)=\frac{4}{3}-\frac{16}{15}=\frac{4}{15}\neq 0$

Should they not return the same value? I also computed the marginal distributions: $$f_X(x)=\int _1^\infty f(x,y)dy=\frac{16}{5x^5}\neq \frac{4}{x^5}$$ $$f_Y(y)=\int _1^\infty f(x,y)dx=\frac{4}{y^6}$$

Is this a mistake in my answer key? When can you use the factoring shortcut to determine if two variables $X,Y$ are independent?

• Is the $f(x,y)$ that you state a valid joint density? May 27, 2017 at 0:43
• I got $\int_1^{\infty}\int_1^{\infty} f(x,y)dxdy = 16/20 \ne 1$. If I am correct, then this exercise has problem. May 27, 2017 at 0:44
• That's what it was. This is not a valid joint density distribution, thanks! May 27, 2017 at 5:27
• Perhaps you could post an answer to your own question? May 27, 2017 at 6:53
• Note that if you were to apply the more general (and careful) formulas modeled after $E(X)=\iint x f(x,y)dxdy / \iint f(x,y) dx dy$, you would have no trouble.
– whuber
May 27, 2017 at 12:52

The answer is that $f(x,y)$ is not a valid joint density distribution because $\int _1^\infty \int _1^\infty f(x,y)dxdy \neq 1$.
• +1 The risk in supplying such an answer is that it does not exhibit any understanding of how to compute covariances, so it might not get you much credit on an exam. To attain high grades, it helps to perceive what the point of a question is and then to fashion a response that constructively addresses that point. In this case, for instance, after pointing out that this density is unnormalized, you should consider going to say that it would be normalized by replacing $16$ by $20$, in which case the covariance would be zero.
• $\int _1^\infty \int _1^\infty \frac{1}{x^5y^6}dxdy=1/20$, so $f(x,y)=\frac{20}{x^5y^6}$ would be the properly normalized joint density distribution. I came to the conclusion today that separating always implies that the variables are independent. $f(x,y)=\frac{4}{x^5}\frac{5}{y^6}$. So, in general, there is no need to check the marginals $f(x)=\int _1^\infty f(x,y)dy=\frac{4}{x^5}$. Let me know if I have the right idea. This is similar to $P[A,B]=P[A]\cdot P[B]$ only when $A,B$ are independent. So, factoring the joint distribution always works as a check for independence. May 29, 2017 at 0:39
• Yes, that's exactly the right idea. See google.com/…. Beware, though, of situations where the domain of $f$ is more complex: if its boundary varies with $x$ or $y$, then the mere factorization of the expression $f(x,y)$ does not suffice as a test of independence. A simple example is the uniform distribution density $f(x,y)=1/\pi$ defined on the unit circle. The key is to write it out as a function defined on all of $\mathbb{R}^2$, using the indicator function $\mathcal{I}$, as $$f(x,y)=\mathcal{I}(x^2+y^2\le 1)\frac{1}{\pi}.$$
• Is this because the marginal densities are $f(x)=\int \frac{1}{\pi }dy=\frac{2}{\pi }\sqrt{1-x^2}$, so that $f(x)f(y)\neq f(x,y)$? Thanks for that indicator function notation, that makes specifying domains much more compact. May 31, 2017 at 2:36