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Math Help - Distribution of Y=min(1,W)

  1. #1
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    Exclamation Distribution of Y=min(1,W)

    Let X be uniformly distributed in (0,1). Define W=2X^(1/2). Find the density of W.
    Let Y= min(1,W). Obtain the distribution of Y.

    Here's what I've done so far:

    P(W<=w) = P(2X^(1/2)<=w) = P(X<= w^2/4) = F(w^2/4) = w^2/4 because X is uniform.

    finally
    f(w) = w/2 0<w<2

    then for the distribution of Y
    P(Y<=y)= P( min(1,W) <=y )

    min(1,W)= 1 if W>1 , W if W<=1 , and 0<min(1,W)<1


    P( 1<=y ) = 0 if 0<y<1 , 1 if y>=1


    and for 0<y<1, P( min(1,W) <=y )= P(1<=y)*P(W>1) + P(W<=1)*P(W<=y) = y^2/16


    so P(min(1,W)<=y) = 0 if y<0 , 1 if y>=1 , y^2/16 if 0<y<1


    I don't know if what I'm doing to obtain y^2/16 is correct, any help is appreciated.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Distribution of Y=min(1,W)

    The relation between the random variables X and Y is...

    \displaystyle Y= \begin{cases} 2\ \sqrt{X} &\phantom{|}\textrm{if } 0 \le X < \frac{1}{4} \\ 1 &\phantom{|}\textrm{if }\frac{1}{4} \le X \le 1 \end{cases} (1)

    ... so that...

    P \{Y \le y\} = \displaystyle \begin{cases} P \{X \le \frac{y^{2}}{4}}\ | X \le \frac{1}{4} \}= y^{2} &\phantom{|}\textrm{if } 0 \le y < 1 \\ 1 &\phantom{|}\textrm{if } y \ge 1 \end{cases}

    ... and the p.d.f. of Y is...

    \displaystyle f(y) = \begin{cases} 2 y &\phantom{|}\textrm{if } 0 \le y < 1 \\ 0 &\phantom{|}\textrm{if } y \ge 1 \end{cases}

    Kind regards

    \chi \sigma
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