1
$\begingroup$

A bit of context: working on a problem about channel coding. Through a channel we are sending a random variable $X_n$, a code, and at the other side we see $Y$ (both discrete). Then we perform an estimation of ML to decide which $X_n$ was sent, but we can make a mistake and detect $X_{n'}$ instead.

Anyway, conditioning to $X$ and $Y$, the probability of an error is

$$ P \left( p(Y|X_{n'})>p(Y|X_n)|X_n,Y \right) $$

To this I want to apply Markov's inequality

$$ \frac{E(A)}{t} \geq P(A>t) $$

But I'm not sure on how should I treat the conditionals. Any help would be appreciated.

$\endgroup$
1
$\begingroup$

Is $p(Y|X_{n'})$ the conditional probability of $Y$ given $X_{n'}$? Then, $p(Y|X_{n'})$ is a non-negative random variable (more precisely, a non-negative measurable function with respect to the sigma-algebra generated by $X_{n'}$). So, if $p(Y|X_n)>0$, we have $$ P \left( p(Y|X_{n'})>p(Y|X_n)\bigg|X_n,Y \right){}\leq{}\dfrac{\mathbb{E}\left[p(Y|X_{n'})\bigg|X_n,Y\right]}{p(Y|X_n)}\,. $$

$\endgroup$
0
$\begingroup$

To elaborate a little on ki3i's answer (+1): if you are given $X_n,Y$, then $p(Y|X_n)=t$ is some constant, and (conditioned on the same) $p(Y | X_{n'})=g(X_{n'})=Z$ is a random variable (function of $X_{n'}$)

Then the probability given is just $P(Z>t) \le \frac{E(Z)}{t}$ Replacing $Z$ by $p(Y | X_{n'}) \mid Y,X_n)$ and $t$ you get the other answer. You can simplify it a little by noting that, in the communication model, the conditioning on $X_n$ in the numerator can be removed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.