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Difficulty understanding ∫ P(X).X^2 dx = <X^2> ?

  1. Oct 13, 2015 #1
    I know for discrete random variables Σ P(x).x = <x>

    Translating for continuous random variables
    I'm also aware of the result ∫ P(x).x dx

    In my lecture notes ( I more or less transcribed from what the lecturer said ):
    ∫ P(x).x^2 dx = <x^2> , should it not be ∫ P(x^2).x^2 dx = <x^2>?

    Does P(x^2) even mean anything in relation to P(x) ? I find it difficult to link the two.

    EDIT: Touching on the ∫ P(x^2).x^2 dx = <x^2> confusion again, for ∫ P(x^2).x^2 dx = <x^2> would you have to be integrating wrt (x^2) too? Ahh, much confusion.

    Could somebody please clear this up for me? Any examples would be much appreciated
  2. jcsd
  3. Oct 13, 2015 #2


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    What is your definition of the notation <x>? Is it the expected value? If so, the expected value of any function, f(x), of x is defined as E(f) =Σ P(x).f(x). That might clarify your remaining questions
    = E(X)
    If your <x^2> notation means the expected value E( X^2 ), then this is true by the definition of expected value.
    No. This must be interpreted as ∫P( X = x2) ⋅ x2 dx. Suppose we have a case where the random variable X is always negative. Then P(X=x2) ≡ 0 for any x. So this integral must be zero.

    For instance, if P(X=-2) = 1 then obviously the expected value of X2 is 4 since X is always -2. This is E(X2) = 4. Your integral would be zero.
  4. Oct 14, 2015 #3


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    Hey Yungphys.

    I think it would help you to think of how to connect the sample mean of [X1 + X2 + ... + Xn]/n to the formula of Sigma p(x)*Xi or Integral p(x)*x*dx and then use a transformation to map the sample mean of [f(X1) + f(X2) + f(X3) + ... + f(Xn))]/n to Sigma*p(x)*f(Xi) or Integral p(x)*x*dx.

    This will make it intuitive.

    To start off thinking about the sample mean formula for a discrete random variable and re-arrange it in terms of p(Xi) = Count(Xi)/n where Count(Xi) adds up the number of times Xi occurs. That will give you the discrete formula and the integral is found by taking appropriate limits.

    Then you can do the same thing for a function of the sample and get the adjusted formula.
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