0
$\begingroup$

What is p(a,b|a) equal to in conditional probability? Any sort of breakdown as to why it might be equal to p(b|a) would be helpful (if it is true in all cases).

My reasoning for asking this question came from a Hastie's (2001) book where one is trying to find E(f(a,b)|a) in the partial dependence plots section. It then got me curious about p(a,b|a).

$\endgroup$
  • $\begingroup$ Try by writing the definition of conditional probability: P(A, B | A) = P(A, A, B) / P(A) = P(A, B) / P(A) = P(B | A). $\endgroup$ – dsaxton Jan 1 at 2:33
1
$\begingroup$

If $a$, $b$ are some events, e.g. $a$: it is raining, $b$: you have your umbrella with you, then, you are asking for P(it is raining and you have umbrella | it is raining). You know that it is raining, so, intuitively, no need to question if it is raining or not again, and this probability equals to P(you have umbrella | it is raining), which will mean $p(a,b|a)=p(b|a)$.

In case of random variables, as @Robin points out, if $A$ and $B$ are the RVs responsible for $a$ and $b$ respectively, the expression should be written in this form: $P(A=a, B=b|A=a)$ in order to prevent any confusions. Because, if $a$ and $b$ are just constants, then what you are asking seems like $p(1,2|1)$, which is ambiguous.

If $a$ and $b$ are random variables themselves, then we lack the particular values. The expression should be like $P(a=a',b=b'|a=a'')$. Here, if $a'' \neq a'$, then given that $a$ equals to $a''$, it cannot be equal to $a'$, and the probability will be $0$. If, $a''=a'$, then again, since we know what the value of $a$ is, the probability will be equal to $p(b=b'|a=a')$, which can be summarized with the help of a $\delta$ function if you like to write it in a compact form, as in @Robin's answer. (P.S. Don't confuse $\delta$ with continuous dirac-delta function).

$\endgroup$
  • $\begingroup$ This is a fantastic response (so is Robin's). Question, are you using the dirac-delta function just to formally extend that you have checked the 0-probability possibilities when a' != a''. Hence, the only non-zero probability (being 1 since it's one option), is when a' = a''. In a way, you had to apply the "cartesian" product of all a' and a'', so to say, to give it a proper analysis. But a compact form would be to say p(A=a,B=b|A=a) results in p(B=b|A=a) for all non-zero probabilities? $\endgroup$ – DoctorDawg Dec 31 '18 at 23:50
  • $\begingroup$ Yes, you understand the $\delta$ function usage, though it is not "dirac-delta". It is more like Indicator function, which gives 1 when inside expression is correct. And, yes p(A=a,B=b|A=a) = p(B=b|A=a) $\endgroup$ – gunes Jan 1 at 12:12
0
$\begingroup$

You have $P(A = a', B = b \mid A = a) = P(B = b \mid A = a) \delta(a = a')$ where $\delta(a = a')$ is equal to 1 only if $a = a'$

$\endgroup$
  • $\begingroup$ Would it instead be equal to p(A,B|A) = p(A|B,A)*p(B|A)? Where I'm getting stuck on is the p(A|B,A) = p(A,B,A)/(p(A,B), which I assume should equal to 1, right? So, what were' left at the end is p(B|A) = p(A,B|A). $\endgroup$ – DoctorDawg Dec 31 '18 at 3:02
  • $\begingroup$ Is this some rule of probability $\endgroup$ – Upendra Pratap Singh Dec 31 '18 at 6:45
  • $\begingroup$ To the best of my knowledge, yes. It is the bayesian decompositions, if you will, of conditional probabilities. $\endgroup$ – DoctorDawg Dec 31 '18 at 23:52

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.