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Useless Data, Conditional Probability, and Independence | Cheenta Probability Series

This concept of independence, conditional probability and information contained always fascinated me. I have thus shared some thoughts upon this.

This concept of independence, conditional probability and information contained always fascinated me. I have thus shared some thoughts upon this.

When do you think some data is useless?

Some data/ information is useless if it has no role in understanding the hypothesis we are interested in.

We are interested in understanding the following problem.

\(X\) is some event. \(Y\) is another event. How much information do \(Y\) and \(X\) give about each other?

We can model an event by a random variable. So, let’s reframe the problem as follows.

\(X\) and \(Y\) are two random variables. How much information do \(Y\) and \(X\) give about each other?

There is something called entropy. But, I will not go into that. Rather I will give a probabilistic view only. The conditional probability marches in here. We have to use the idea that we have used the information of \(Y\), i.e. conditioned on \(Y\). Hence, we will see how \(X \mid Y\) will behave?

How does \( X \mid Y\) behave? If \(Y\) has any effect on \(X\), then \(X \mid Y\) would have changed right?

But, if \(Y\) has no effect on \(X\), then \(X \mid Y\) will not change and remain same as X. Mathematically, it means

\( X \mid Y\) ~ \(X\) \(\iff\) \( X \perp \!\!\! \perp Y\)

We cannot distinguish between the initial and the final even after conditioning on \(Y\).

Theorem

\(X\) and \(Y\) are independent \( \iff \) \( f(x,y) = P(X =x \mid Y = y) \) is only a function of \(x\).

Proof

\( \Rightarrow\)

\(X\) and \(Y\) are independent \( \Rightarrow \) \( f(x,y) = P(X =x \mid Y = y) = P(X = x)\) is only a function of \(x\).

\( \Leftarrow \)

Let \( \Omega \) be the support of \(Y\).

\( P(X =x \mid Y = y) = g(x) \Rightarrow \)

\( P(X=x) = \int_{\Omega} P(X =x \mid Y = y).P(Y = y)dy \)

\(= g(x) \int_{\Omega} P(Y = y)dy = g(x) = P(X =x \mid Y = y) \)

Exercises

  1. \((X,Y)\) is a bivariate standard normal with \( \rho = 0.5\) then \( 2X – Y \perp \!\!\! \perp Y\).
  2. \(X, Y, V, W\) are independent standard normal, then \( \frac{VX + WY}{\sqrt{V^2+W^2}} \perp \!\!\! \perp (V,W) \).

Random Thoughts (?)

How to quantify the amount of information contained by a random variable in another random variable?

Information contained in \(X\) = Entropy of a random variable \(H(X)\) is defined by \( H(X) = E(-log(P(X)) \).

Now define the information of \(Y\) contained in \(X\) as \(\mid H(X) – H(X|Y) \mid\).

Thus, it turns out that \(H(X) – H(X|Y) = E_{(X,Y)} (log(\frac{P(X \mid Y)}{P(X)})) = H(Y) – H(Y|X) = D(X,Y)\).

\(D(X,Y)\) = Amount of information contained in \(X\) and \(Y\) about each other.

Exercise

  • Prove that \(H(X) \geq H(f(X))\).
  • Prove that \(X \perp \!\!\! \perp Y \Rightarrow D(X,Y) = 0\).

Note: This is just a mental construction I did, and I am not sure of the existence of the measure of this information contained in literature. But, I hope I have been able to share some statistical wisdom with you. But I believe this is a natural construction, given the properties are satisfied. It will be helpful, if you get hold of some existing literature and share it to me in the comments.

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