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Let x and y be random variables whose probabilities depend on an unknown parameter θ. I am specifically interested in the case that both x and y are Bernoulli, but the question below can also be asked in general. Assuming sufficient regularity, the Fisher information of a set X of random variables about θ is given by the expression:

IX(θ)=E[2θ2logf(X;θ)|θ],

where f(X;θ) is the probability density of X conditional on θ.

If x and y are independent, then:

Ix,y(θ)=Ix(θ)+Iy(θ).

Question: Is it always true (even when x and y are dependent) that Ix,y(θ)Ix(θ)+Iy(θ)?

My intuition from Shannon information tells me that surely this is so- if you have a coin, and you toss it once, your optimal strategy for the second toss is surely just to toss it again. But I've been unable to find a proof for this, and the analysis gets very complicated very fast when I try to prove this "subadditivity" property analytically.

Question: Do you have a reference for this property?

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You should take a look at Stam's proof which is the original proof of Fisher information sub-addivity, and maybe on this one too which uses a generalized result. This paper contains a proof of a related result. Hope it helps!

-- edit You're right, this is another property, sorry for being misleading. According to my PhD advisor, it does not hold; sometimes super-additivity, (other inequalities).

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  • Thanks! Is this the correct subadditivity property though? It looks to me like subadditivity with respect to something else- namely IT(x,y)Ix,y for any function T, where x,y are iid (equation 1.5 in Stam). Have I missed something? Commented Jun 11, 2015 at 21:42
  • I've skimmed through all three references you gave now, and can't find the statement of subadditivity in the sense that I need it- could you point me to it? Note in particular that a Bernoulli random variable does not have a continuous probability density function (of course); the references seem to all be working with condituous pdfs. Commented Jun 12, 2015 at 6:09
  • Thank you for this. I now see that the opposite sometimes holds (because the joint holds more information than the sum of the marginals)- but I am still pretty sure that we have subadditivity in the Bernoulli case. Thinking further... Commented Jun 12, 2015 at 13:20

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