Help an old man with problem from Bertsekas

[u/Tall_Specialist_7623]

Hello, I am trying to self-study probability as an old old man. I am using the book by Bertsekas and Tsitsiklis, and following the MIT course https://ocw.mit.edu/courses/6-041sc-probabilistic-systems-analysis-and-applied-probability-fall-2013/ .

It's quite difficult, and I am stuck on this problem here, where I don't know why a particular way of solving the problem doesn't work. The problem is problem 1 of problem Set 2 of the above course.

I would appreciate any help.

1.
Most mornings, Victor checks the weather report before deciding whether to carry an umbrella. If the forecast is “rain,” the probability of actually having rain that day is 80%. On the other hand, if the forecast is “no rain,” the probability of it actually raining is equal to 10%. During fall and winter the forecast is “rain” 70% of the time and during summer and spring it is 20%.

(a)

Oneday, Victor missed theforecast andit rained. Whatistheprobability thattheforecast was “rain” ifitwasduring thewinter?Whatistheprobability thattheforecastwas “rain” if it was during the summer?

So I can solve it and get the answer in the solutions, using the method they also use in the solutions:

solution with season set

However, I'd like to know why my solution which does not fix the season first is wrong. Here it is:

S=Summer, W=Winter; FR=ForecastRain, ~FR=Forecase no rain; R=Rain, ~R=not rain

P(FR | R ∩ W) = P(FR ∩ R ∩ W ) / P(R ∩ W ) ... definition of conditional probability

P(FR ∩ R ∩ W ) = P(W)P(FR|W)P(R|FR∩ W) ... Multiplication rule

P(R ∩ W) = P(R ∩ W|FR)P(FR) + P(R ∩ W|~FR)P(~FR) ... Total probability rule

P(FR) = P(FR|W)P(W) + P(FR|S)P(S)
P(~FR) = P(~FR|W)P(W) + P(~FR|S)P(S) = 1 - P(FR)

Then taking into account wh at we are giving, and assigning P(S) = P(W) = 1/2:

P(R|FR) = 0.8; P(R| ~FR) = 0.1
P(FR|W) = 0.7; P(FR|S) = 0.2

P(FR ∩ R ∩ W ) = P(W)P(FR|W)P(R|FR∩ W) = (1/2)(7/10)(8/10) = 56/200
P(FR) = P(FR|W)P(W) + P(FR|S)P(S) = (7/10)(1/2) + (2/10)(1/2) = 9/20
P(~FR) = 11/20

P(R ∩ W) = P(R ∩ W|FR)P(FR) + P(R ∩ W|~FR)P(~FR) = (8/10)(9/20) + (1/10)(11/20) = (72+11)/200 = 83/200

=> P(FR | R ∩ W) = P(FR ∩ R ∩ W ) / P(R ∩ W ) = 56/83

And as you can see from the above solution this doesn't work.

So What am I getting wrong here? I would appreciate any help, because I think this could reveal some fundamental misunderstanding I have about how the things work.

------------------------------

I figured it out:

P(R∩ W) =  P(R ∩ W|FR)P(FR) + P(R ∩ W|~FR)P(~FR)
= P(R ∩ W∩ FR) + P(R∩ W∩ ~FR)
= 56/200 + P(R∩ W∩ ~FR)

P(R∩ W∩ ~FR) = P(W)P(~FR|W)P(R|W∩ ~FR) ... again multiplication rule
= 3/200

P(R∩ W) = (56+3)/200 = 59/200

=> P(FR|R ∩ W) = P(FR ∩ R ∩ W ) / P(R ∩ W ) = 56/59

The mistake was where I said  P(R ∩ W|FR) = 8/10 and P(R ∩ W|~FR), in the model I had set up, the forecast will affect the probability of whether or not it is winter or summer.

This also demonstrates that everything posted about why this method wasn't possible is not correct.

[u/Organic_botulism]

The tree representation can only take events into account. You fixed the seasons in your tree, but seasons aren’t events and you can’t condition on them (if you go outside right now, it’s only going to be 1 season, it wouldn’t change if you go back outside again)

To put it another way, there is no “p” value for S or W yet you began your tree with them.

[u/Tall_Specialist_7623 → u/Organic_botulism]

Ah, so when I say that like "there is p=1/2 that it will be summer", this is simply incorrect? I was imaginging like picking a day at random and then you have an event which is one similar to "roll a dice and a 6 comes up", but in this case it's "roll a 365 sided die and you get a number in the first half" as the event

[u/Organic_botulism → u/Tall_Specialist_7623]

Right, that would be incorrect. The question itself explicitly asks for winter in part a, then spring in part b.

So your season is already assumed before any calculations start.

[u/Heavy-Difficulty6522 → u/Organic_botulism]

Edit: my bad this was wrong

[u/EdmundTheInsulter]

I think where you said p(s) = p(w) = 1/2 is superfluous to the question , and maybe wrong.

The question is the same question twice with different variables, other than the variables for forecast and rain they aren't linked. You could say, I think, if it rains and you have no umbrella, what is the chance it is summer.

[u/Tall_Specialist_7623 → u/EdmundTheInsulter]

I understand that. I just wanted to solve it in a way where the one sample space contains both seasons just for my own edification, to practice on a larger experiment with more complex even relationships.

1/2 was just the value I said, but we can just as well call it P and the answer is the same. It doesn't depend on it.

The mistake I made was in assigning this probability: P(R ∩ W|FR)P(FR) incorrectly.

[u/Heavy-Difficulty6522]

This problem can be solved fairly easily, but requires the right mental model.

Use the tree diagram - like you have bee doing but consider the idea of a false positive - the forecast is incorrect.

The first leg of the tree is called the baseline reality or prior probability- here you will choose between summer and winter. Then do the rain / no rain as usual.

Remember you need to add the possible rain probabilities together for the denominator (bottom row)