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    Probability based on multiple pieces of information




    It's been about 20 year since my last stats class and I'm working on a fun little side project that involves some probability. I'm having some trouble remembering all my stats.

    Let's say I have a bird feeder that feeds 1 bird at a time. From past observations I know that if there is a bird at the bird feeder, there is a 50% probability that the bird is a Cardinal, 30% it's a Blue Jay, and 20% it's a Raven.

    However, I can also detect it's wing span. Let's say a Cardinal has an average wing span of 5.7" and a standard deviation of 0.2". A Blue Jay has an average wing span of 5.9" and a standard deviation of 0.15". A Raven has an average wing span of 6.3" and a standard deviation of 0.25".

    If there is a bird at the feeder with a wing span of 5.8", what are the odds it is a Cardinal, Blue Jay, or Raven?

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    Re: Probability based on multiple pieces of information

    Sounds like you want to use Bayes Theorem.
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    Re: Probability based on multiple pieces of information

    How could you use means and SD's with BT?

    I certainly don't know a lot about BT, but I thought that it deals with the probabilities of a discrete number of outcomes (e.g., beak type, or tail color), whereas wingspan is a continuous variable.
    Last edited by bruin; 07-21-2014 at 01:27 AM.

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    Re: Probability based on multiple pieces of information

    Quote Originally Posted by bruin View Post
    How could you use means and SD's with BT?

    I certainly don't know a lot about BT, but I thought that it deals with the probabilities of a discrete number of outcomes (e.g., beak type, or tail color), whereas wingspan is a continuous variable.
    You can turn the event with respect to that continuous variable into a discrete one. Make event A be the event that the bird's wing span is 5.8" +/- 1". We actually want to ignore the +/- 1" part since that was fairly arbitrary. We'll deal with that later but for the time being what we're interested in is P(Bird Type | A).

    Now let the +/- 1" get smaller (so say +/- 1/12" ... then 1/1000"). Now P(A | Bird Type) will basically converge to using the density value for the distributions in the bayes theorem calculations as we let the width of the interval get smaller and smaller.
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    Re: Probability based on multiple pieces of information

    I guess I still don't get it...how can you have a density value for a normally distributed variable equaling a specific value, like 5.8?

    If A were "wingspan < 5.8" or "wingspan > 5.8" I would get how to find P(A|Bird Type), but for A is "wingspan = 5.8" I don't get how to find P(A|Bird Type).

    Can you maybe give an example with numbers of how, in the process of using BT to calculate P(Raven|A), one would calculate the needed value to plug in for P(A|Raven)?

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    Re: Probability based on multiple pieces of information

    Well the argument I'm making is that you just use the density from the pdf. I'm justifying that this is alright (since it's not truly a probability) through the use of calculating probabilities that are essentially "the probability of observing a bird with a wing span within a very small neighborhood of 5.8 inches". If keep making that neighborhood smaller and smaller then everything ends up being proportional to just using the density in place of the actual probability calculation. Too tired to do an example with actual numbers at the moment.
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    Re: Probability based on multiple pieces of information

    I was doing some reading on Bayes Theorem last night and that did seem like what I wanted to use, but like Bruin said, I couldn't figure out how to use standard deviation to express the probability a bird is a Raven if the wing span is 5.8". That's where I think I'm stuck.

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    Re: Probability based on multiple pieces of information

    You'll probably need to make an assumption about the distribution of the wing spans for each of the birds. It would probably be reasonable to assume a normal distribution.
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    Re: Probability based on multiple pieces of information

    Yes I'm assuming a normal distribution.

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    Re: Probability based on multiple pieces of information


    Do you know how to get density values for a normal distribution?
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