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Probability measures

  1. Sep 7, 2005 #1
    Do Someone know reply this question? It seems easy, but I am not being able to resolve it.

    Show that if P and Q are two probability measures defined on the same (countable) sample space, then a.P + b.Q is also a probability measure for any two nonnegative numbers a and b satisfying a + b = 1. Give a concrete illustration of such a mixture
  2. jcsd
  3. Sep 7, 2005 #2


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    The basic approach is to look at the definition of what a probability measure is. Then you go through and check if the mixture as defined in the question satisfies those conditions given that P and Q are already probability measures.
  4. Sep 7, 2005 #3
    Probability measure

    David, I know that an probability measure on same space [itex]\Omega[/itex] is a function of subsets of [itex]\Omega[/itex] satisfying three axioms:

    (i) For every set [itex]A \subset \Omega[/itex], the value of the function is a non-negative number: P(A) [itex]\geqslant[/itex] 0.

    (ii) For any two disjoint sets A and B, the value of the function for their union A + B is equal to the sum of its value for A and its value for B:

    P(A + B) = P(A) + P(B) provided A.B = [itex]{\O}[/itex].

    (iii) The value of the function for [itex]\Omega[/itex] (as a subset) is equal to 1:

    P([itex]\Omega[/itex]) = 1.

    but I don't know how to show that a.P + b.Q is too.
  5. Sep 7, 2005 #4


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    Perhaps you're not sure of the process foor how to show the axioms apply to the new measure so I'll give you an example with the first one.

    Remember that the a.P + b.Q is a function that maps a subset A onto a real number. You already know some properties of how the functions P and Q work and you use them to work out the properties of the combined function.

    So for the first axiom, we want to show that a.P + b.Q (which I'll call R for the moment) is non-negative. i.e. that R(A) >=0 for any A. The definition of P and Q say that P(A)>=0 and Q(A)>=0. Also, from the conditions set in the question, a and b are >=0. So for any A, the function R(A) = a.P(A) + b.Q(A) is a non-negative times a non-negative plus a non-negative times a non-negative. This total is non-negative, which is what you need for the axiom to hold.

    What I just did there might seem overkill, but explicitly remembering that you can do this for a particular subset A and then realizing that it works for any subset A will sometimes get you to the answer you need. That will be helpful to organize yout thoughts for axiom ii. Axiom iii works just like the proof of axiom i.
  6. Sep 8, 2005 #5

    Thank you for help. I responsed the question with success.
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