# Understanding probability in Updating information sequentially

I am reading book "First course in Probability" by Sheldon Ross.

Under example with title "Updating information sequentially", he says following:

Question
Suppose there are $n$ mutually exclusive and exhaustive possible hypotheses, with initial (sometimes referred to as prior) probabilities $P(H_i), \sum_{i=1}^n=1$. Now, if information that the event $E$ has occurred is received, then the conditional probability that $H_i$ is the true hypothesis (sometimes referred to as the updated or posterior probability of $H_i$) is $$P(H_i|E)=\frac{P(E|H_i)P(H_i)}{\sum_jP(E|H_j)P(H_j)} ...Equation(I)$$ If the information that events $E_1$ and $E_2$ occurred is received, then the conditional probability is given as: $$P(H_i|E_1E_2)=\frac{P(E_1E_2|H_i)P(H_i)}{\sum_jP(E_1E_2|H_j)P(H_j)}$$ One might wonder, however, when one can compute $P(H_i|E_1E_2)$ by using the right side of Equation $(I)$ with $E=E_2$ and with $P(H_j)$ replaced by $P(H_j|E_1)$, $j = 1, . . . , n$. That is, when is it legitimate to regard $P(H_j|E_1), j>=1$, as the prior probabilities and then use equation $(I)$ to compute the posterior probabilities?

In solution, the author says following:

Solution

1. The answer is that the preceding is legitimate, provided that, for each j = 1, . . . , n, the events E1 and E2 are conditionally independent

2. Author then proves: $$P(H_i|E_1E_2)=\frac{P(E_2|H_i)P(H_i|E_1)}{\sum_{i=1}^nP(E_2|H_i)P(H_i|E_1)}$$

3. Finally author concludes the solution as follows: For instance, suppose that one of two coins is chosen to be flipped. Let $H_i$ be the event that coin $i$, $i = 1, 2,$ is chosen, and suppose that when coin $i$ is flipped, it lands on heads with probability $p_i, i = 1, 2$. Then the preceding equations show that, to sequentially update the probability that coin 1 is the one being flipped, given the results of the previous flips, all that must be saved after each new flip is the conditional probability that coin 1 is the coin being used. That is, it is not necessary to keep track of all earlier results.

I am really struggling to wrap my head around this as I am not able to make much sense out of this. Also I am struggling to form any concrete question to ask here. But still I have formed some to help me make it some sense. But anyways any general more clear explanation is welcome. So the questions are:

1. The last sentence in the question start with "That is,". How the last two sentences of the question equivalent? That is, how putting $E=E_2$ and $P(H_j)=P(H_j|E_1)$ means $P(H_j|E_1)$ is the prior probability and $E_2$ is the posterior probabilities?
2. Does the last two sentences in the question starting with "One might wonder,..." mean to say $P(H_i|E_1E_2)=P(E_2|H_iE_1)$
3. How does the title "Updating information sequentially" makes sense?
4. How does the last/third point in the solution makes sense?

We can imagine doing a sequence of experiments $E_1,$ $E_2,\ldots$ with binary outcomes where the event $E_i$ indicates whether one of the outcomes happened. These experiments naturally affect your view on the probability each of the hypotheses $H_i$ is true. We have prior probabilities $P_0(H_i)$ before any of the experiments are done. Then we can write updated probabilities $P_1(H_i)$ for after the first experiment has been down, $P_2(H_i)$ for after the second, etc. And we know from Bayesian probability what these are: $$P_k(H_i) = P(H_i|E_1,E_2,\ldots,E_k)= \frac{P(E_1,\ldots,E_k|H_i)P(H_i)}{\sum_j P(E_1,\ldots,E_k|H_j)P(H_j)}$$

Okay, so you have information if $E_1$ happened or not, so you update the probability of hypothesis $H_i$ as $$P_1(H_i) = \frac{P(E_1| H_j)P_0(H_j)}{\sum_j P(E_1|H_j)}...$$ that one's easy. The question the author is asking is as follows:

When can we compute $P_2(H_i)$ as $$P_2(H_i) = \frac{P(E_2|H_i)P_1(H_i)}{\sum_j P(E_2|H_j)P_1(H_j)}$$ and $P_3(H_i)$ as $$P_3(H_i)=\frac{P(E_3|H_i)P_2(H_i)}{\sum_j P(E_3|H_j)P_2(H_j)}$$ and so on?

What the author is imagining is before each new experiment you have a "running prior" and then the new posterior computed using Bayes on the new experiment $E_k$ and the aforementioned "prior" $P_{k-1}(H_i)$ becomes the next value of the "running prior". Essentially what the author wants to do is to at each round forget everything about the previous experiment except what we got for the posterior and then just treat that as a prior for the next round.

Now a little scrutiny tells us that this isn't always possible. There's a lot more information in the results of the previous experiments than just how they made you feel about the hypotheses. In particular the detailed pattern of results could drastically alter how we feel about the following results. Silly example: say there's one hypothesis $H$ (and its complement) that is "this animal has cancer" but the animal could be a mouse or an elephant. Say the (reasonable) first experiment was you check whether the specimen is a mouse or an elephant. Now, elephants might have a lower or higher natural incidence of cancer than mice (I'm guessing lower) so you will update your probability here. However, imagine if that's all the information you carried along with you to the next round and you forgot whether or not the animal was a mouse or an elephant. That would probably be information you'd like to have to interpret the other experiments.

As the author says, the issue is whether the experiments are conditionally independent of one another given the hypothesis. In the preceding case it's implausible that the results of previous experience (given the presence or nonpresence of cancer) are independent of whether or not the animal is a mouse or an elephant.

But in other cases conditional independence might hold, like if the experiments are flipping a coin and the hypotheses are statements about the coin's bias. In this case we can actually just keep a "running prior" as in the author's plan. We can show that this works for $P_2(H_i)$ as follows: recall we have $$P_1(H_i) = \frac{P(E_1|H_i)P_0(H_i)}{\sum_j P(E_1|H_j)P_0(H_j)}$$ and $$P_2(H_i) = \frac{P(E_1,E_2|H_i)P_0(H_i)}{\sum_j P(E_1,E_2|H_j)P_0(H_j)}.$$ We can then compute $$\frac{P(E_2|H_i)P_1(H_i)}{\sum_k P(E_2|H_k)P_1(H_k)} \\= \frac{P(E_2|H_i)\frac{P(E_1|H_i)P_0(H_i)}{\sum_j{P(E_1|H_j)P_0(H_j)}}}{\sum_k P(E_2|H_k)\frac{P(E_1|H_k)P_0(H_k)}{\sum_j P(E_1|H_j)P_0(H_j)}} \\= \frac{P(E_2|H_i)P(E_1|H_i)P_0(H_i)}{\sum_k P(E_2|H_k)P(E_1|H_k)P_0(H_k)} \\= \frac{P(E_1,E_2|H_i)P_0(H_i)}{\sum_j P(E_1,E_2|H_j)P_0(H_j)} \\= P_2(H_i)$$ where at the end we used conditional independence: $P(E_1,E_2|H_i) = P(E_1|H_i)P(E_2|H_i).$

1. The last sentence in the question start with "That is,". How the last two sentences of the question equivalent? That is, how putting $E=E_2$ and $P(H_j)=P(H_j|E_1)$ means $P(H_j|E_1)$ is the prior probability and $E_2$ is the posterior probabilities?
2. Does the last two sentences in the question starting with "One might wonder,..." mean to say $P(H_i|E_1E_2)=P(E_2|H_iE_1)$

They're just wondering if there is an analogous form, and what is required.   Turns out the answers are "yes" and "when the events are conditionally independent when given a hypothesis".

\def\P{\operatorname{\sf P}} \begin{align} \P(H_i\mid E) & = \dfrac{\P(E\mid H_i)\P(H_i)}{\sum_j \P(E\mid H_j)\P(H_j)} \\[1ex] \P(H_i\mid E_2,E_1) & = \dfrac{\P(E_2\mid H_i, E_1)\P(H_i\mid E_1)}{\sum_j \P(E_2\mid H_j,E_1)\P(H_j\mid E_1)} \\[1ex] & = \dfrac{\P(E_2\mid H_i)\P(H_i\mid E_1)}{\sum_j \P(E_2\mid H_j)\P(H_j\mid E_1)} &E_1\perp E_2\mid H_i \end{align}

1. How does the title "Updating information sequentially" makes sense?

If you have calculated all the conditional probabilities for hypothesis given events, $\P(H_i\mid E_1)$ you can now update when given additional information by the next event in the sequence, $\P(H_i\mid E_1, E_2)$.

1. How does the last/third point in the solution makes sense?

Suppose I have two coins with biased probability for obtaining heads, $p_1, p_2$, and I randomly select a coin and toss it, obtaining a heads. $E_1$ being the event of doing so.

$$\P(H_i)=\tfrac 12, \P(E_1\mid H_i)=p_i$$

I can calculate $\P(H_1\mid E_1)$ and $\P(H_2\mid E_1)$

\begin{align}\P(H_i\mid E_1) & = \dfrac{p_i\cdotp\tfrac12}{p_1\cdotp\tfrac12+p_2\cdotp\tfrac12} \\ & = \dfrac{p_i}{p_1+p_2}\end{align}

I toss the same coin once more, obtaining another head, and $E_2$ is the event of doing so. $$\text{again }\P(E_2\mid H_i)=p_i$$ Now, I could calculate $\P(H_1\mid E_1,E_2)$ from scratch, but since the events are conditionally independent given a particular coin, I may update the previous calculation.

\begin{align}\P(H_i\mid E_1,E_2) & = \dfrac{p_i \cdot \dfrac{p_i}{p_1+p_2}}{p_1 \cdot \dfrac{p_1}{p_1+p_2}+p_2 \cdot \dfrac{p_2}{p_1+p_2}} \\ & = \dfrac{p_i^2}{p_1^2+p_2^2}\end{align}

(In this scenario it would have been easier to find from scratch, but that is not always going to be the case.)

• (1) What did you meant by $E_1⊥E_2∣H_i$. Especially by $⊥$? Is it $P(E_2|H_i,E_1)=P(E_2|H_i)$ because $E_1$ and $E_2$ are independent? (2) Do you mean $E_1E_2$ (or $E_1\cap E_2$) by $E_1,E_2$? (I guess obviously yes, just that never saw comma separated notation in books) Also whats more common / standard? Or its just personal taste?
– RajS
Aug 15, 2017 at 9:59
• (1) $E_1\perp E_2\mid H_i$ means $E_1,E_2$ are conditionally independent given $H_i$. (2) Lists in probability functions are conjunctive, so yes, it is $E_1\cap E_2$. You should have encountered it; it is a standard notation (albeit more often used with random variables than events). Aug 15, 2017 at 12:10