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Can a random variable donimate

  1. Jul 22, 2010 #1
    Given two random variables x and y, and a constant c

    What conditions are needed to make:

    [tex]Prob( w x + y < c ) \approx Prob( w x < c ), \text{ for } w \rightarrow \infty[/tex]

    Can anyone help? I think [tex]E(x) < \infty[/tex] and [tex]E(y) < \infty[/tex] might do. Is this right?

    tks!!!!
     
    Last edited: Jul 22, 2010
  2. jcsd
  3. Aug 3, 2010 #2
    I think you need the expected value of y an order of magnitude less than E(x).
     
  4. Aug 4, 2010 #3
    In the limit as [tex] w \rightarrow \infty [/tex] I believe they are always equal. I will use the probability density functions (f(x),f(y), and f(x,y)) to give my reasoning.

    [tex] P(wx < c) = P(x < c/w) = P(x < 0)[/tex] in the limit of [tex]w \rightarrow \infty[/tex]
    [tex] = \int_{-\infty}^{0}f(x)dx = \int_{-\infty}^{0}\int_{-\infty}^{\infty}f(x,y)dydx [/tex]

    To calculate the probability you have to add up the region of the density for which wx+y < c, which can be achieved by integrating for each x from y=-infinity to the line y=-wx+c:

    [tex] P(wx +y < c) = \int_{-\infty}^{\infty}\int_{-\infty}^{-wx+c}f(x,y)dydx [/tex]

    In the limit this becomes the y-axis, so we actually have in this case:
    [tex] = \int_{-\infty}^{0}\int_{-\infty}^{\infty}f(x,y)dydx = \int_{-\infty}^{0}f(x)dx[/tex]

    and so they are equivalent.
     
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