0
$\begingroup$

I hope you don't mind it is a little bit long. Actually it is not very complicated derivation.

Suppose $X\sim B(n, p)$, $Y \sim B(m, q) $, and 0 < p, q < 1. Both X and Y are either 0 or 1.

$p = \mathrm{P}(X=1|Y=1)$ and $q = \mathrm{P}(X=1|Y=0)$

According to Bayes theorem, I derive below equation, I label it as equation (1)

\begin{equation} \begin{split} log \frac{\mathrm{P}(Y=1|X)}{\mathrm{P}(Y=0|X)} &= log \frac{\mathrm{P}(X|Y=1)\mathrm{P}(Y=1)}{\mathrm{P}(X|(Y=0)\mathrm{P}(Y=0)} \\&=log \frac{\mathrm{P}(X|Y=1)}{\mathrm{P}(X|(Y=0)}+log\frac{\mathrm{P}(Y=1)}{\mathrm{P}(Y=0)} \end{split} \end{equation}

I represent $\mathrm{P}(X|Y=1)$ as $\mathrm{P}(X|Y=1) = X(\mathrm{P}(X=1|Y=1)) + (1-X)\left(1-\mathrm{P}\left(X=1|Y=1\right)\right)$ (formula 1)

Reasoning 1:

My reasoning about the formula 1 is that, since X is either 0 or 1, so when X=1, $\mathrm{P}(X|Y=1) = X(\mathrm{P}(X=1|Y=1))$, and when X=0, $\mathrm{P}(X|Y=1) = (1-X)(1-\mathrm{P}(X=1|Y=1))$. combining this two situation, I obtained $\mathrm{P}(X|Y=1) = X(\mathrm{P}(X=1|Y=1)) + (1-X)\left(1-\mathrm{P}\left(X=1|Y=1\right)\right)$. Whether my reasoning in this step is problematic or not?

If this step is correct, then with the same reasoning I obtained $\mathrm{P}(X|Y=0) = X(\mathrm{P}(X=1|Y=0)) + (1-X)\left(1-\mathrm{P}\left(X=1|Y=0\right)\right)$ (formula 2)

Recall equation (1), I substitute the formulas 1 and 2 in the equation (1), it gives,

\begin{equation} \begin{split} log \frac{\mathrm{P}(Y=1|X)}{\mathrm{P}(Y=0|X)} &= log \frac{\mathrm{P}(X|Y=1)\mathrm{P}(Y=1)}{\mathrm{P}(X|(Y=0)\mathrm{P}(Y=0)} \\&=log \frac{\mathrm{P}(X|Y=1)}{\mathrm{P}(X|(Y=0)}+log\frac{\mathrm{P}(Y=1)}{\mathrm{P}(Y=0)} \\&= log\frac{X(\mathrm{P}(X=1|Y=1)) + (1-X)\left(1-\mathrm{P}\left(X=1|Y=1\right)\right)}{X(\mathrm{P}(X=1|Y=0)) + (1-X)\left(1-\mathrm{P}\left(X=1|Y=0\right)\right)}+log\frac{\mathrm{P}(Y=1)}{\mathrm{P}(Y=0)} \end{split} \end{equation}

Recall $p = \mathrm{P}(X=1|Y=1)$ and $q = \mathrm{P}(X=1|Y=0)$, equation (1) could be represent as \begin{equation} \begin{split} log \frac{\mathrm{P}(Y=1|X)}{\mathrm{P}(Y=0|X)} &= log\frac{Xp + (1-X)\left(1-p\right)}{Xq + (1-X)\left(1-q\right)}+log\frac{\mathrm{P}(Y=1)}{\mathrm{P}(Y=0)} \end{split} \end{equation}

Reasoning 2:

For the expression $log\frac{Xp + (1-X)\left(1-p\right)}{Xq + (1-X)\left(1-q\right)}$, if X =1 , it equals to $log\frac{p }{q }$, if X =0, it equals to $log\frac{1-p }{1-q }$, so the expression $log\frac{Xp + (1-X)\left(1-p\right)}{Xq + (1-X)\left(1-q\right)}$ could be represented as $log\frac{Xp + (1-X)\left(1-p\right)}{Xq + (1-X)\left(1-q\right)} = X log\frac{p }{q } + (1-X)(log\frac{1-p }{1-q }) = X (log\frac{p }{q } -log\frac{1-p }{1-q }) + log\frac{1-p }{1-q }$. Is my reasoning in the step correct? Can I derive this expression in this way?

Finally, if my above dervation is correct, equation (1) could be represent as: \begin{equation} \begin{split} log \frac{\mathrm{P}(Y=1|X)}{\mathrm{P}(Y=0|X)} &= X (log\frac{p }{q } -log\frac{1-p }{1-q }) + log\frac{1-p }{1-q }+log\frac{\mathrm{P}(Y=1)}{\mathrm{P}(Y=0)} \\&= X (log\frac{p(1-q) }{(1-p)q }) + log\frac{(1-p)\mathrm{P}(Y=1)}{(1-q)\mathrm{P}(Y=0) } \end{split} \end{equation}

I am a little bit uncertain for two part of reasoning. It looks reasonable, but not very rigorous, so I want to ask for some advices. My question have been labelled in boldface. Could somebody help to check for me? Can I derive in this way when the variable is binary.

$\endgroup$
0
$\begingroup$

Your reasoning looks correct as long as $X,Y$ are Bernoulli random variables, and not generally Binomial as aluded to in the opening line.

To be strict, you should be using indicator random variables, $\mathbf 1_{X=1}$, $\mathbf 1_{X=0}$ rather that $X$ itself.   However, that is a technicality when indeed $~X=\mathbf 1_{X=1}~$ and $~(1-X) = \mathbf 1_{X=0}~$.

$\endgroup$

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.