# What would p(a,b|a) be equal to?

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).

• 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). – dsaxton Jan 1 at 2:33

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).

• 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? – DoctorDawg Dec 31 '18 at 23:50
• 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) – gunes Jan 1 at 12:12

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 $$a = a'$$

• 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). – DoctorDawg Dec 31 '18 at 3:02
• Is this some rule of probability – Upendra Pratap Singh Dec 31 '18 at 6:45
• To the best of my knowledge, yes. It is the bayesian decompositions, if you will, of conditional probabilities. – DoctorDawg Dec 31 '18 at 23:52