# Distribution of Y=min(1,W)

• June 17th 2011, 09:02 PM
LAINHELL
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.
• June 18th 2011, 06:24 AM
chisigma
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

$\chi$ $\sigma$