- #26

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Huh, I'm using that too along with 6.9 and 3.45, but I can't get that answerYes, but with the 0.6 factor included, and using the correct choices of ##EX_1## and ##EX_2##.

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- Thread starter Rifscape
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- #26

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Huh, I'm using that too along with 6.9 and 3.45, but I can't get that answerYes, but with the 0.6 factor included, and using the correct choices of ##EX_1## and ##EX_2##.

- #27

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Is the number you got 0.4539?

- #28

Ray Vickson

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Is the number you got 0.4539?

My p is (1-your p), so the answers agree when expressed in words.

Sorry for the confusion: I have been suffering from a cold that makes my head fuzzy, and so have been a bit mixed up.

- #29

StoneTemplePython

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To be honest,

First criticism: see prior page where I pointed out that the problem is illogical if questions are during a given week for use in the following week, if we are talking about Stephane coming up with questions during the last week of the course. Based on results, it seems that the renewal idea is out the door, so my criticism that the question is busted is sustained. The fix would be to say that Stephane makes the decision to 'reward' students at the beginning of the second to last week of the course (so that the questions can be issued during the last week).

The equation your professor wants is, have a prior distribution = ##\left[\begin{matrix}0.55\\0.45\end{matrix}\right]## and then use

##posterior \propto diag(likelihood) \left[\begin{matrix}0.55\\0.45\end{matrix}\right]

##

and your prof wants you to assume that 5 questions were issued -- and use that for your likelihood function.

##posterior \propto

\left[\begin{matrix}p(5, \mu= 3.45) & 0\\0 & p(5, \mu= 6.9)\end{matrix}\right]

\left[\begin{matrix}0.55\\0.45\end{matrix}\right]

##

##posterior \propto

\left[\begin{matrix}

0.12929992

& 0\\0 &

0.13135067

\end{matrix}\right] \left[\begin{matrix}0.55\\0.45\end{matrix}\right]

##

if your normalize the above (i.e. make sure the resulting vector sums to one), you get

##posterior =

\left[\begin{matrix}0.546102377853564\\0.453897622146436\end{matrix}\right]

##

- - - -

The issue is this is

Stéphane decides to "reward" the students by no longer limiting himself to 4 exercises, and instead assigning every exercise he writes.

or, more cumbersomely, it could say:

Stéphane decides to "reward" the students by no longer limiting himself to 4 exercises, and instead assigning every exercise he writes.

but what it actually says (using my top student setup) is:

Stéphane decides to "reward" the students by no longer limiting himself to 4 exercises, and instead assigning every exercise he writes.

In this setup, the random variable is actually the number of questions on the test, not the probability of having a visitor -- the probability of a visitor is just a parameter we are estimating.

So the statement we actually get is far more general, and different, than being told that the top student has seen the homework assignment and that the top student expects to get 5 correct, and hence that there are only 5 questions being assigned (i.e. the number of questions being assigned is no longer a random variable).

The issue is that:

##posterior^T \left[\begin{matrix}3.45 \\ 6.9 \end{matrix}\right] =

\left[\begin{matrix}0.546102377853564\\0.453897622146436\end{matrix}\right]

^T \left[\begin{matrix}3.45 \\ 6.9 \end{matrix}\right] =

5.0159467964052062 \neq 5 ##

Hence the solution vector violates the law of iterated expectations, unless the student has seen the number of homework questions and determined it equals 5.

- #30

Ray Vickson

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@Rifscape

To be honest,part (c) is probably one of the worst worded questions that I've seen. I hope you have a higher quality text to study from or is reviewing something like MIT's 6.041 (https://ocw.mit.edu/courses/electri...s-analysis-and-applied-probability-fall-2010/ ) or Harvard's intro to probability (see Joe Blitzstein on youtube). Otherwise there's a risk of learning the opposite of clear thinking from this course.

Let me simplify, and consider a simpler but probabilistically identical problem where instead of a student with a 60% chance of answering an exercise is considered, consider the case where top student with 100% chance of answering an exercise correctly is expected to answer 5 questions correctly.

It makes perfectly good sense to ask for the posterior probability of a visiting scholar, given that a student answered 3 questions correctly and

First of all, if 3 questions are answered correctly the test must contain ##N \geq 3## questions. For event ##\bar{S}## we have ##N \sim \text{Poisson}(\alpha)## with ##\alpha = 6.9,## while for event ##S## we have ##N \sim \text{Poisson}(\beta),## with ##\beta = 3.45##. Given an ##N = n \geq 3## the probability the student answers 3 questions correctly is the Binomial probability ##C(n,3) p^3 q^{n-3}, ## where ##p = 0.6, q = 0.4##. So, if ##Y## is the number of correctly-answered questions, then for event ##\bar{S}## we have ##P_{\bar{S}}(Y=3\: \&\: N=n) = C(n,3) p^3 q^{n-3} \alpha^n e^{-\alpha}/n!,## which simplifies to

$$P_{\bar{S}}(Y=3\: \&\: N=n) = \frac{p^3 \alpha^3}{3!} \frac{e^{-\alpha} (\alpha q)^{n-3}}{(n-3)!}.$$

Thus,

$$P(Y = 3|\bar{S}) = \sum_{n=3}^{\infty} P_{\bar{S}}(Y = 3\: \& \:N = n) = \frac{(\alpha p)^3}{3!} e^{-\alpha + \alpha q} = \frac{(\alpha p)^3 e^{-\alpha p}}{3!}.$$

Of course, this is Poisson probability with mean ##\alpha p = 0.6 \times 6.9 = 4.14.## Similarly, for event ##S##, ##Y## is Poisson with mean ##p \beta = 0.6 \times 3.45 = 2.07##. We have ##P(Y=3|\bar{S}) = 4.14^3 e^{-4.14}/3! \doteq 0.1883,## while ##P(Y=3|S) = 2.07^3 e^{-2.07}/3! \doteq 0.1865.##

The posterior probability of no visiting scholar, given that ##\{ Y = 3 \}##, is

$$P(\bar{S}|Y=3) = \frac{P(Y=3|\bar{S}) P(\bar{S})}{P(Y=3)} = \frac{(0.45)(0.1883)}{(0.45)(0.1883) + (0.55)(0.1865)}.$$

- #31

StoneTemplePython

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It makes perfectly good sense to ask for the posterior probability of a visiting scholar,given that a student answered 3 questions correctlyandwithout making any assumption that the test has 5 questions. Admittedly the problem is more challenging than the "assume 5 questions" version, and may possibly be beyond the ability/knowledge of the OP, but I really do not know what the instructor intended, so having two possible versions cannot hurt...

First of all, if 3 questions are answered correctly the test must contain ##N \geq 3## questions.$$

I think your response is fair and you are definitely right that the wording is open for multiple interpretations. In my book, having an observation is a very big deal and the wording is silent whether or not the student has even received the assignment... all we have to go on is, that the student "is expected to answer 3 questions correctly." It seems to me that clarity is important generally in math and especially so in probability, but this question is not clear.

That said, maybe OP's prof meant to write a good homework question, but was interrupted by a visitor from Switzerland.

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