document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1170932>Question 1170932</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let Y = X^2 where X is U(a,b), 0<=a<b.  \r<BR>\n" );
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document.write( "Show that the pdf of Y is f(y) - 1/[2sqrt(y)(b-a)] for a^2<=y <= b^2.\r<BR>\n" );
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document.write( "I know to start I need x = sqrt(y) and to plug into the formula f(g(y))du/dy to find the pdf but I am struggling from there. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851061 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=851061\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's derive the probability density function (PDF) of $Y = X^2$, where $X$ is uniformly distributed on the interval $[a, b]$, with $0 \le a < b$.\r<BR>\n" );
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document.write( "**1. PDF of X**\r<BR>\n" );
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document.write( "Since $X$ is uniformly distributed on $[a, b]$, its PDF is:\r<BR>\n" );
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document.write( "$$f_X(x) = \begin{cases} \frac{1}{b-a} & \text{for } a \le x \le b \\ 0 & \text{otherwise} \end{cases}$$\r<BR>\n" );
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document.write( "**2. Cumulative Distribution Function (CDF) of Y**\r<BR>\n" );
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document.write( "We want to find $F_Y(y) = P(Y \le y)$. Since $Y = X^2$, we have:\r<BR>\n" );
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document.write( "$$F_Y(y) = P(X^2 \le y) = P(-\sqrt{y} \le X \le \sqrt{y})$$\r<BR>\n" );
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document.write( "Since $a \ge 0$, $X$ is non-negative, so we only need to consider the positive root:\r<BR>\n" );
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document.write( "$$F_Y(y) = P(a \le X \le \sqrt{y})$$\r<BR>\n" );
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document.write( "Now, we can express this in terms of the CDF of $X$:\r<BR>\n" );
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document.write( "$$F_Y(y) = F_X(\sqrt{y}) - F_X(a)$$\r<BR>\n" );
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document.write( "The CDF of $X$ is:\r<BR>\n" );
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document.write( "$$F_X(x) = \begin{cases} 0 & \text{for } x < a \\ \frac{x-a}{b-a} & \text{for } a \le x \le b \\ 1 & \text{for } x > b \end{cases}$$\r<BR>\n" );
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document.write( "So,\r<BR>\n" );
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document.write( "$$F_Y(y) = \frac{\sqrt{y} - a}{b-a} - \frac{a-a}{b-a} = \frac{\sqrt{y} - a}{b-a}$$\r<BR>\n" );
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document.write( "This is valid for $a^2 \le y \le b^2$.\r<BR>\n" );
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document.write( "**3. PDF of Y**\r<BR>\n" );
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document.write( "To find the PDF of $Y$, we differentiate the CDF of $Y$ with respect to $y$:\r<BR>\n" );
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document.write( "$$f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy} \left( \frac{\sqrt{y} - a}{b-a} \right)$$\r<BR>\n" );
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document.write( "$$f_Y(y) = \frac{1}{b-a} \frac{d}{dy} (\sqrt{y} - a) = \frac{1}{b-a} \frac{d}{dy} (y^{1/2} - a)$$\r<BR>\n" );
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document.write( "$$f_Y(y) = \frac{1}{b-a} \left( \frac{1}{2} y^{-1/2} - 0 \right) = \frac{1}{2(b-a)\sqrt{y}}$$\r<BR>\n" );
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document.write( "Therefore, the PDF of $Y$ is:\r<BR>\n" );
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document.write( "$$f_Y(y) = \frac{1}{2\sqrt{y}(b-a)} \quad \text{for } a^2 \le y \le b^2$$\r<BR>\n" );
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document.write( "And $f_Y(y) = 0$ otherwise.<BR>\n" );
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