+ Reply to Thread
Results 1 to 2 of 2

Thread: Normal Average versus Weighted Average

  1. #1
    Points: 14, Level: 1
    Level completed: 27%, Points required for next Level: 36

    Location
    Plymouth, MA
    Posts
    1
    Thanks
    0
    Thanked 0 Times in 0 Posts

    Normal Average versus Weighted Average




    I'm taking a self-study course from "The Great Courses" called "Joy of Numbers". I'm on the lesson "Joy of Mathematical Games" and am confused by the professor's seemingly interchangeable use of the terms "weighted probabilities" and "weighted averages".

    The example given is the Law of Total Probability where:

    If event B has 3 possible outcomes B1, B2, B3, then

    P(A) = P(A|B1)P(B1)+P(A|B2)P(B2)+P(A|B3)P(B3)

    The professor calls the result the "Weighted Average" of the event's probabilities.

    So, if this is an average, why isn't everything divided by 3?

    e.g. P(A) = (P(A|B1)P(B1)+P(A|B2)P(B2)+P(A|B3)P(B3))/3

    When I work through a real life scenario like -
    A horse (A|B) wins a race 60% of the time when it rains, 50% of the time when it snows and 20% of the time when it's clear. And, the probability of rain (B) is 80%, snow is 19% and clear skies is 1%, then the percentage probability of the horse winning any given race is:

    P(horse win) = (0.60)(0.80)+(0.50)(0.19)+(0.20)(0.01) = 0.577

    Intuitively that makes sense because you would expect that with a high probability of rain the horse would come close to 60% probability of winning.

    But I'm having trouble reconciling the term "average" with something that doesn't take the number of terms in the equation into consideration.

    Can anyone help me with this?

  2. #2
    TS Contributor
    Points: 12,287, Level: 72
    Level completed: 60%, Points required for next Level: 163
    rogojel's Avatar
    Location
    I work in Europe, live in Hungary
    Posts
    1,471
    Thanks
    160
    Thanked 332 Times in 312 Posts

    Re: Normal Average versus Weighted Average


    hi,
    a normal average with 3 terms can be thought of as x1*1/3+x2*1/3+x3*1/3. A general weighted average will be something like x1*a1+x2*a2+x3*a3 with the condition that a1+a2+a3=1.

    In your case, because P(B1)+P(B2)+P(B3)=1 this is a weighted average .

    I hope this helps

    regards

+ Reply to Thread

           




Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats