10
$\begingroup$

Let's say I'm trying to find out the probability that someone's favorite ice cream flavor is vanilla.

I know that the person also enjoys horror movies.

I want to find out the probability that the person's favorite ice cream is vanilla given that they enjoy horror movies.

I know the following things:

  1. $5\%$ of people choose vanilla as their favorite ice cream flavor. ( This is my $P(A)$ )
  2. $10\%$ of people whose favorite is vanilla ice cream also love horror movies. ( This is my $P(B|A)$ )
  3. $1\%$ of people whose favorite is not vanilla ice cream also love horror movies ( This is my $P(B|\lnot A)$ )

So, I calculate it like this: $$P(A|B)=\frac{0.05\times0.1}{(0.05 \times 0.1)+(0.01 \times(1-0.05))}$$ I find that $P(A|B) = 0.3448$ (rounded to the nearest ten-thousandth). There is a $34.48\%$ chance that a horror movie fan's favorite ice cream flavor is vanilla.

But then I learn that the person has seen a horror movie in the past 30 days. Here's what I know:

  1. $34.48\%$ is the updated posterior probability that vanilla is the person's favorite ice cream flavor -- the $P(A)$ in this next problem.
  2. $20\%$ of people whose favorite is vanilla ice cream have seen a horror movie in the past 30 days.
  3. $5\%$ of people whose favorite is not vanilla ice cream have seen a horror movie in the past 30 days.

This gives: $$\frac{0.3448\times0.2}{(0.3448\times0.2)+(0.05\times(1-0.3448))} = 0.6779$$ when rounded.

So now I believe there is a $67.79\%$ chance that the horror movie fan loves ice cream given that they've seen a horror movie in the past 30 days.

But wait, there is another thing. I also learned that the person owns a cat.

Here's what I know:

  1. $67.79\%$ is the updated posterior probability that vanilla is the person's favorite ice cream flavor -- the $P(A)$ in this next problem
  2. $40\%$ of people whose favorite is vanilla ice cream also own cats
  3. $10\%$ of people whose favorite is not vanilla ice cream also own cats

This gives: $$\frac{0.6779\times0.4}{(0.6779\times 0.4)+(0.1\times(1-0.6779))} = 0.8938$$ when rounded.

My question basically boils down to this: Am I correctly updating probability using Bayes' theorem? Am I getting anything else wrong in my methods?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ love = favorite? you're not posting degrees of loving. if you love it, it is your favorite. clarify if needed. $\endgroup$
    – generic_user
    Dec 19, 2012 at 9:10
  • $\begingroup$ Good point. I changed "love" to "favorite." It's not grammatically correct, but it's less wordy than saying "choose vanilla for their favorite ice cream flavor." I hope that clears things up. $\endgroup$
    – user1626730
    Dec 19, 2012 at 14:16

1 Answer 1

Reset to default
7
$\begingroup$

This is not correct. Sequential updating of this type only works when the information you are receiving sequentially is independent (e.g. iid observations of a random variable). If each observation is not independent, as in this case, you need to consider the joint probability distribution. The correct way to update would be to go back to the prior, find the joint probability that someone loves horror movies, has seen a horror movie in the last 30 days, and owns a cat given that they do or do not choose vanilla as their favorite ice cream flavor, and then update in a single step.

Updating sequentially like this when your data are not independent will rapidly drive your posterior probability much higher or lower than it ought to be.

Improve this answer
$\endgroup$
13
  • 1
    $\begingroup$ How do you mean by "when the information you are receiving sequentially is independent?" If you mean "independent of the event you're trying to predict," do you know how I can tell if the info I'm getting is independent? $\endgroup$
    – user1626730
    Dec 19, 2012 at 15:19
  • $\begingroup$ Conditionally independent given the event you are trying to predict. If they were independent of the event you're trying to predict then they wouldn't do you any good. As for how you can tell--you have to think about what your data is. In this case, whether someone has watched a horror film in the last 30 days is clearly not independent of whether they love horror films. $\endgroup$
    – Jonathan Christensen
    Dec 19, 2012 at 15:32
  • $\begingroup$ When you say "conditionally independent," I'm guessing you mean that each P(B) (i.e., horror-movie-loving, cat-ownership) aren't related to one another? If so, wouldn't the cat-ownership variable be independent of the horror-movie-loving? $\endgroup$
    – user1626730
    Dec 19, 2012 at 15:47
  • $\begingroup$ Yes, you can make an argument that cat-ownership is independent of horror-movie-loving. It's not, necessarily, though--e.g., maybe women are both more likely to love cats and less likely to love horror movies. $\endgroup$
    – Jonathan Christensen
    Dec 19, 2012 at 16:18
  • $\begingroup$ Hm, I'm not quite sure what you mean by adding in that bit about women and cats. Could you explain further, please? $\endgroup$
    – user1626730
    Dec 19, 2012 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.