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Assume that $V_i,V_j,D$ are all dependent random variables and real-valued, and let the matrix $H$ be defined by

$$ H_{ij} = \mathrm E(V_i V_j) - \mathrm E(\mathrm E(V_i\mid D)\mathrm E(V_j\mid D)) $$

My goal is to determine whether $H$ is positive semi-definite, it is the Hessian of a log-likelihood function that I would like to know whether it is convex.

If I can show that $\langle V_i, V_j \rangle = H_{ij}$ is an inner product then $H$ is Gramian and so it is positive semi-definite.

The first two axioms for an inner-product follows directly

$$ \langle V_i, V_j \rangle = \langle V_j, V_i \rangle $$ $$ \langle aV_i, V_j \rangle = a \langle V_j, V_i \rangle $$

Now, to determine whether the third axiom holds

$$ \langle V_i, V_i \rangle \geq 0 $$

I need to determine whether it is true that

$$ \mathrm E(V_i^2) - \mathrm E(\mathrm E(V_i\mid D)^2) \geq 0 $$

$$ \mathrm E(V_i^2) - \mathrm E(V_i)^2 \geq 0 $$

$$ Var(V_i) \geq 0 $$

From Schwarz' inequality $\mathrm E(V_i)^2 \leq \mathrm E(V_i^2)$, so $\langle V_i, V_i \rangle \geq 0$ is true.

I am quite far from my comfort zone. Is my reasoning OK?

Update: Would also be interesting to hear about other ways that one can prove $H_{ij}$ is positive-semi-definite.

share|cite|improve this question
For SEMI-definite, you don't need to show that $Var(V_i)=0$ off $V_i=0$. – Fabian Jan 24 '12 at 23:22
@Fabian, thanks. – j-a Jan 25 '12 at 7:56
up vote 0 down vote accepted

I still can't see anything wrong with the approach in the question.

I figured out an alternative way to prove it:

$$ H_{ij} = \mathrm E(V_i V_j) - \mathrm E(\mathrm E(V_i\mid D)\mathrm E(V_j\mid D)) $$

Noting that $$ \begin{align} Cov(X,Y|Z) & = E[(X-E[X|Z])(Y-E[Y|Z])] \\ & = E[XY] - E[X|Z]E[Y|Z] \end{align} $$

We get

$$ H_{ij} = E[Cov(V_i,V_j|Z)] $$


$$ A_{ij} = Cov(V_i,V_j|Z) $$

Which is positive semidefinite

$$ x^TAx \geq 0 \quad \forall x $$

also $$ E[x^TAx] =x^TE[A]x = x^THx \geq 0 \quad \forall x $$

So $H$ is positive semidefinite.

share|cite|improve this answer
If anyone else answers the original question, I will accept that as the answer instead. – j-a Jan 28 '12 at 10:01

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