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I came across this question in Probability & Statistics by Rohatgi Exercise 1.6.9.

Let $P(A|B) = P(A|B\cap C)P(C) + P(A|B\cap C^c)P(C^c)$ with $P(A|B\cap C) \neq P(A|B)$.

If all the three events have non-zero probability, prove that $B$ and $C$ are independent.

After several efforts, I couldn't solve this seemingly innocuous looking question. Any idea on how to proceed will be appreciated.

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Since ${\bf P}[B], {\bf P}[A\,\vert\,B]>0$ then, using the given equation, note that we can relate two ways of computing $\,{\bf P}[A \cap B]$:

$${\bf P}[A \,\vert\, B \cap C ]\,{\bf P}[ B \cap C ] + {\bf P}[ A \,\vert\, B \cap C^c ]\,{\bf P}[ B \cap C^c ] \\= {\bf P}[A \cap B ] \\= {\bf P}[A \,\vert\, B \cap C ]\,{\bf P}[B]{\bf P}[C] + {\bf P}[ A \,\vert\, B \cap C^c ]\,{\bf P}[B]\,{\bf P}[C^c]\tag{1}$$

$(1)$ holds non-trivially so that we can subtract the r.h.s from the l.h.s, resulting in

$$\bigg({\bf P}[ B \cap C ]-{\bf P}[ B ]\,{\bf P}[ C ]\bigg)\bigg({\bf P}[A \,\vert\, B \cap C ] - {\bf P}[ A \,\vert\, B \cap C^c ]\bigg)=0\tag{2}$$

Above, for the factorisation, we have used the fact that ${\bf P}[C] > 0$. Now, we know that ${\bf P}[A \,\vert\, B \cap C ] \neq {\bf P}[ A \,\vert\, B \cap C^c ]$ since, otherwise, the given equation in the question implies $ {\bf P}[A \,\vert\, B ] = {\bf P}[ A \,\vert\, B \cap C ]$; a contradiction. Hence, equation $(2)$ implies

$${\bf P}[ B \cap C ]={\bf P}[ B ]\,{\bf P}[ C ]\,.\tag{3}$$

All other probabilities of relevant event combinations - such as ${\bf P}[B\cap C^c] = {\bf P}[B]{\bf P}[C^c]$ - follow by using $(3)$ with basic relationships between probabilities of non-null events and sub-events.

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  • $\begingroup$ Thanks. Very elegantly explained. $\endgroup$ – user2808118 Dec 13 '15 at 15:24
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HINT

Proof by contradiction: Assume that events $B$ and $C$ are dependent events.

Therefore we can say that:

$$P(B \cap C) = P(B) + P(C) - P(B \cup C)$$

Now substitute this in your formula above and if the equality doesn't hold, then the latter is true and $B$ and $C$ are independent events.

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    $\begingroup$ The equality mentioned by you is true whether or not the events are independent. $\endgroup$ – user2808118 Dec 12 '15 at 14:47

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