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I have the following example:

There are two coins, labeled 1 and 2, either or both of which are possibly biased. The probability of a head is $$P(H \mid \text{coin} \ i) = p_i, \ \ \ \ (i = 1, 2).$$ Choose a coin at random and toss it twice. Define events $A_j = \{ \text{$H$ occurs on the $j^{\text{th}}$ toss} \} \ (j = 1, 2)$, and $B = \{ \text{coin 1 is chosen} \}$. Then
(i) $A_1$ and $A_2$ are not independent unless $p_1 = p_2$.
(ii) Given $B$, $A_1$ and $A_2$ are independent events.

I'm having trouble understanding this. Why would we require $p_1 = p_2$ in order for $A_1$ and $A_2$ to be independent? After all, if $H$ occurs on one toss, then I don't see how that would affect $H$ occurring for the other toss. Furthermore, how does $B$ change anything? I don't see how the event $B$ changes anything for $A_j$ at all.

EDIT: For the record, it's possible that this example is incorrect, so there's no obligation to justify its correctness.

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2 Answers 2

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I believe the statements are correct. Keep in mind that independence is defined through probabilities, so those are up for inspection.

If two events $A_1$ and $A_2$ are independent, then $P(A_1) = P(A_1|A_2)$, meaning that if you knew that the second coin toss resulted in heads, it will not affect the probability of the first coin toss being heads. Or in other words, the probability of the first toss being heads is the same whether the second toss is heads or tails $P(A_1|A_2) = P(A_1|\bar{A_2})$.

In this example, the event $A_1$ can occur in two ways: coin 1 is chosen and the first toss results in heads, or coin 2 is chosen and the first toss results in heads, and we have that $$P(A_1) = P(B)\cdot P(A_1|B) + P(\bar{B}) \cdot P(A_1|\bar{B})$$ $$P(A_1|B) = p_1, \,\,P(A_1|\bar{B}) = p_2.$$

(ii) Just states that if you toss the same coin twice, the probability of the outcome of the first toss is unaltered by the outcome of the second toss, and vice versa. However, since your two coins do not necessarily have the same probability for heads, it does matter which coin you toss.

(i) To see that this point is valid, it can be useful to look at extremes. Let $p_1>0$ and $p_2= 0$. If the second toss results in heads, then you can be certain that you are tossing coin 1, and hence $P(A_1) = p_1$ since you now know that $P(B) = 1$ and $P(\bar{B}) = 0$, so clearly occurence of $A_2$ gives information about $P(A_1)$. For $p_1 = p_2$, the outcome of the second coin toss does not provide any information regarding which coin was tossed, but for $p_1 \neq p_2$, $P(B)$ can be updated accordingly.

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Note: please edit this answer if you can make it more intuitive, or if any corrections are needed.

For independence we need $Pr(A_1 \cap A_2) = Pr(A_1) Pr(A_2)$.

The probabilities of events $A_i$ depend on $B$, so we start with $Pr(A_1 \cap A_2 | B)$.

$Pr(A_1 \cap A_2 | B) = Pr(A_1 | A_2 \cap B) Pr(A_2 | B) = Pr(A_1 | B) Pr(A_2 | B)$.

The last equality is because the two coin tosses are independent once a $B$ has been selected.

To remove the dependence of the event B we require the probabilities of the coins to be the same. $Pr(A_1 | B) = $Pr(A_1) when the probabilities of the two coins are the same because then the coins are identical, and whichever is selected will yield the same probabilities of returning a head. The same applies to $A_2$. So we have that $Pr(A_1 \cap A_2 | B) = Pr(A_1) Pr(A_2)$. Since the RHS does not depend in $B$ the LHS cannot either and hence we get the independence definition.

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  • $\begingroup$ Thanks for the answer. Can you please explain how the probabilities of events $A_i$ depend on $B$? Perhaps I'm misinterpreting the example. $\endgroup$ Mar 21, 2021 at 9:40
  • $\begingroup$ Suppose coin one is biased one way and coin two is biased the other way. For the first toss, $A_1$, will depend on event $B$ which is whether the first coin is chosen. $\endgroup$ Mar 21, 2021 at 9:47
  • $\begingroup$ But the event $A_1$ is just whether $H$ occurs on the 1st toss, so I don't see how this depends on the event that coin 1 is chosen. For instance, just because the event $B$ occurs and coin 1 is chosen doesn't mean anything for whether $H$ occurs on the $j = 1$ or $j = 2$ toss. In other words, it seems that $A$ and $B$ are independent, no? $\endgroup$ Mar 21, 2021 at 9:56
  • $\begingroup$ For the record, it's possible that this example is incorrect, so I'm not trying to present it as absolutely correct. $\endgroup$ Mar 21, 2021 at 10:04
  • $\begingroup$ Conditional probability has a temporal element to it, which I think makes it more complicated. When do events $A_i$ and $B$ occur in time? Event $B$ has to occur first, and this is how it affects $A$. So although at the time of $A_i$, $B$ is a variate and not a random variable, it ($B$) still provided randomness to the whole process at a given point (when a coin was selected). $\endgroup$ Mar 21, 2021 at 10:13

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