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Math Help - Distribution to density to expected

  1. #1
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    Distribution to density to expected

    A random variable X has the cumulative distribution function
    F(x) =   0 for x<1
    F(x) =  \frac{x^2 - 2x + 2}{2} for 1 \leq x < 2
    F(x) = 1  for x \geq 2

    Calculate the variance of X.

    The answer specifies that the density function is
    f(x) = 0.5  if x=1
    f(x) = x-1  if 1<x<2
    f(x) = 0    otherwise

    Then
    E(X) = 0.5 + \int_1^2 x(x-1) dx
    E(X^2) = 0.5 + \int_1^2 x^2(x-1) dx

    I got the f(x) = x-1 part, and I got how to calculate the variance after you have the expected values, but I'm lost on other questions.

    My questions are:
    Where do we get f(x) = 0.5  if x=1? F(1) = 0.5, but I can't figure out why f(1) would equal 0.5.

    What is the rule for putting parts of the stepwise density function into the expected value equations? I don't know what the rule is called so I don't know how to review it. We're adding the slope at a single point to the slope over a big area, which is something I can't quite work out visually.
    Last edited by Boris B; April 24th 2008 at 02:16 PM.
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  2. #2
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    Quote Originally Posted by Boris B View Post
    A random variable X has the cumulative distribution function
    F(x) = 0 for x<1
    F(x) = \frac{x^2 - 2x + 2}{2} for 1 \leq x < 2
    F(x) = 1 for x \geq 2

    Calculate the variance of X.

    The answer specifies that the density fucntion is
    f(x) = 0.5 if x=1
    f(x) = x-1 if 1<x<2
    f(x) = 0 otherwise

    Then
    E(X) = 0.5 + \int_1^2 x(x-1) dx
    E(X^2) = 0.5 + \int_1^2 x^2(x-1) dx

    I got the f(x) = x-1 part, and I got how to calculate the variance after you have the expected values, but I'm lost on other questions.

    My questions are:
    Where do we get f(x) = 0.5 if x=1? F(1) = 0.5, but I can't figure out why f(1) would equal 0.5.

    What is the rule for putting parts of the stepwise density function into the expected value equations? I don't know what the rule is called so I don't know how to review it. We're adding the slope at a single point to the slope over a big area, which is something I can't quite work out visually.
    F(x) = 0 for x<1 implies that \Pr(X < 1) = 0.

    F(x) = \frac{x^2 - 2x + 2}{2} for 1 \leq x < 2 implies that \Pr(X \leq 1) = \frac{1^2 - 2(1) + 2}{2} = \frac{1}{2}.

    It follows that \Pr(X = 1) =  \Pr(X \leq 1) - \Pr(X < 1) = \frac{1}{2}.
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  3. #3
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    Quote Originally Posted by Boris B View Post
    A random variable X has the cumulative distribution function
    F(x) = 0 for x<1
    F(x) = \frac{x^2 - 2x + 2}{2} for 1 \leq x < 2
    F(x) = 1 for x \geq 2

    Calculate the variance of X.

    The answer specifies that the density fucntion is
    f(x) = 0.5 if x=1
    f(x) = x-1 if 1<x<2
    f(x) = 0 otherwise

    Then
    E(X) = 0.5 + \int_1^2 x(x-1) dx
    E(X^2) = 0.5 + \int_1^2 x^2(x-1) dx

    I got the f(x) = x-1 part, and I got how to calculate the variance after you have the expected values, but I'm lost on other questions.

    My questions are:
    Where do we get f(x) = 0.5 if x=1? F(1) = 0.5, but I can't figure out why f(1) would equal 0.5.
    The 1/2 comes from the jump discontinuity in the cumulative distribution F(x) at x=1, indicating a probability mass of 1/2 at that point.

    Another way of thinking about this is to think of the density as a generalised function, then we may represent the density of a piecewise continuous cumulative distribution as the sum of a continuous function and delta functionals at the discontiuities of amplitude equal to the size of the jumps.

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Boris B View Post
    A random variable X has the cumulative distribution function
    F(x) = 0 for x<1
    F(x) = \frac{x^2 - 2x + 2}{2} for 1 \leq x < 2
    F(x) = 1 for x \geq 2

    Calculate the variance of X.

    The answer specifies that the density fucntion is
    f(x) = 0.5 if x=1
    f(x) = x-1 if 1<x<2
    f(x) = 0 otherwise
    If this last is the given answer for the "density" then it is wrong, as as given it is not a density since:

    \int_{-\infty}^{\infty} f(x)~dx=1/2

    The density is the generalised function:

    f(x)=g(x)+(1/2)\delta(x-1)

    where

    g(x) = x-1, \ \ 1<x<2.
    g(x) = 0, \ \ \mbox{otherwise}.

    RonL
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Boris B View Post

    What is the rule for putting parts of the stepwise density function into the expected value equations? I don't know what the rule is called so I don't know how to review it. We're adding the slope at a single point to the slope over a big area, which is something I can't quite work out visually.
    If you use the generalised function approach you have a perfectly normal looking equation for the expected value. The problem arrises if you represent the distribution as mixed discrete/continuous, when you have to work with the two components seperately.

    RonL
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