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Thread: Distribution function for a function of random variable

  1. April 8th 2010, 10:35 AM #1
    tttcomrader
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    Distribution function for a function of random variable

    Suppose that X is a random variable with distribution function F_X(t) and let Y=a+bX \ \ \ b<0 . Derive the distribution function for Y.

    My solution:

    I know that F_Y(t)=P(T \leq t)=P(a+bX \leq t)

    =P(X \geq \frac {t-a} {b} ) =1 - P( X \leq \frac {t-a}{b})=1- F_X ( \frac {t-a}{b} )

    But the answer is wrong, why?

    Thanks.
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  2. April 8th 2010, 11:27 AM #2
    Moo
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    Hello,

    More precisely, it's 1-\mathbb{P}\left(X{\color{red}<} \frac{t-a}{b}\right)

    There are some situations (in particular for some discrete random variables) where \mathbb{P}\left(X< \frac{t-a}{b}\right)\neq \mathbb{P}\left(X\leq \frac{t-a}{b}\right)
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  3. April 8th 2010, 12:12 PM #3
    tttcomrader
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    For some reason, the back of the book says the answer is:

    1-F_X( \frac {t-a}{b})+p_X ( \frac {t-a}{b})
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  4. April 8th 2010, 12:14 PM #4
    Anonymous1
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    Quote Originally Posted by tttcomrader View Post
    For some reason, the back of the book says the answer is:

    1-F_X( \frac {t-a}{b})+p_X ( \frac {t-a}{b})
    So that confirms what Moo suggested...

    You have 1 - P( X < \frac {t-a}{b}) So you just need to add the case P( X = \frac {t-a}{b})
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