I am a new beginner in stats. I have specifically diverted my attention towards this because, I wish to understand the concept of Deep Bayesian Learning, so I am starting with the basics. The question is:

The Bayes rule equation is given by

$P(X | Y)$ = $\cfrac{P(Y | X). P(X)}{P(Y)}$

But, I have noticed in some places, the denominator being ignored entirely and using just the numerator of the RHS of the equation. making it:

$P(X | Y)$ = $P(Y|X).P(X)$

Is there some special case where we can ignore the P(Y)? as in when P(Y) = 1? But, if that's the case, wouldn't all the things become very easy: P(Y | X) will become 1, and P(X | Y) will be just P(X) and done.

  • $\begingroup$ A reference to "some places" might be helpful $\endgroup$ – Juho Kokkala Feb 19 '18 at 6:12
  • $\begingroup$ Instead of $P(X|Y)=P(Y|X)P(X)$ did you intended to write $P(X|Y) \propto P(Y|X)P(X)$? The latter form is correct while the former is not, since it is not always a proper probability. $\endgroup$ – Sycorax Feb 19 '18 at 6:15
  • $\begingroup$ @JuhoKokkala , I was refering to this video -> youtube.com/watch?v=o2Tpws5C2Eg $\endgroup$ – Animesh Karnewar Feb 19 '18 at 6:20
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    $\begingroup$ It's horrible video, I'm sorry, ridden with errors $\endgroup$ – Aksakal Feb 19 '18 at 7:01
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    $\begingroup$ @AnimeshKarnewar Just want to second Aksakal. That video is garbage, and what he writes is just wrong. $\endgroup$ – Matthew Drury Feb 19 '18 at 7:10

Indeed, ignoring P(Y) is very common. This is not done since P(X) equals one. This is done when we are interested which value of X is more probable, like in classification.

For positive, $P(X=Positive | Y)$ = $\cfrac{P(Y | X=Positive)* P(X=Positive)}{P(Y)}$

For negative, $P(X=Negative | Y)$ = $\cfrac{P(Y | X=Negative)* P(X=Positive)}{P(Y)}$

If we want to know if $P(X=Positive | Y) > P(X=Negative | Y)$, we can omit ${P(Y)}$ from both expressions. ${P(Y)}$ is a probability and therefore it is a positive constant and can be removed.

Think of a medical example. A person Y goes into a physician office that would like to know if Y suffers from X. ${P(Y)}$ is the probability the Y will step into the office. Y might be very common or rare but it doesn't matter, he is already in the office.

Add to that, that measuring ${P(Y)}$ is no longer a medical question. The physician knowledge cannot help with that. Therefore we prefer avoiding coping with an unnecessary hard problem.

Note that if you would like to be precise, after omitting P(Y) you should write $P(X=Positive | Y) \approx P(Y | X=Positive)* P(X=Positive)$ and use approximation instead of equality.

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    $\begingroup$ well, in your last equation, $P(Y)$ is not really omitted. Moreover, in your example, $P(Y)$ would be the probability of the characteristics/variables that predicts $X$, not the probability of going into the office. For example, if $X=\text{suffers from cancer}$, $P(Y)$ would be the probability that the blood test is positive (unless the probability of having cancer depends on you going into the doctor's office, which of course is absurd). Notice that you used $X$ for the dependent variable (has cancer?) and Y for the independent one $\endgroup$ – raffaem Apr 27 '19 at 9:33
  • $\begingroup$ Thanks. I should have omit the P(Y) there. In that part I meant to say that if you omit you should say it is an approximation and not equality. I hope it is clearer now. As for the rest, it is just Bayes rule.See it as knowledge given event and not as causality. $\endgroup$ – DaL Apr 28 '19 at 5:49

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