document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209942>Question 1209942</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Given the random variables 𝑋, 𝑌, 𝑍 with the following distributions<BR>\n" );
document.write( "𝑋 ~ 𝒩(0, 2^2), 𝑌 ~ 𝒰(0, |𝑋|), 𝑍 ~ ℰ(1/𝑌)<BR>\n" );
document.write( "where: 𝒩(𝜇, 𝜎^2) is the normal distribution with mean 𝜇 and variance 𝜎^2, 𝒰(𝑎, 𝑏) is the uniform distribution on the interval (𝑎, 𝑏), ℰ(𝜆) is the exponential distribution with mean 1/𝜆.<BR>\n" );
document.write( "Calculate 𝑃(𝑋 ≥ 3) given that we observe 𝑍 = 4. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851145 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2189)</B></A><img src=\"/images/stars/stars-13.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=851145\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's break down this problem step by step.\r<BR>\n" );
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document.write( "**1. Understand the Distributions**\r<BR>\n" );
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document.write( "* **X ~ N(0, 2^2):** X is normally distributed with mean 0 and variance 4 (standard deviation 2).<BR>\n" );
document.write( "* **Y ~ U(0, |X|):** Y is uniformly distributed on the interval (0, |X|). This means the value of Y depends on the absolute value of X.<BR>\n" );
document.write( "* **Z ~ E(1/Y):** Z is exponentially distributed with mean Y.\r<BR>\n" );
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document.write( "**2. Probability Density Functions (PDFs)**\r<BR>\n" );
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document.write( "* **X:**<BR>\n" );
document.write( "    * $f_X(x) = \frac{1}{\sqrt{2\pi(2^2)}} e^{-\frac{(x-0)^2}{2(2^2)}} = \frac{1}{2\sqrt{\pi}} e^{-\frac{x^2}{8}}$<BR>\n" );
document.write( "* **Y | X = x:**<BR>\n" );
document.write( "    * $f_{Y|X}(y|x) = \frac{1}{|x|}$ for $0 < y < |x|$, and 0 otherwise.<BR>\n" );
document.write( "* **Z | Y = y:**<BR>\n" );
document.write( "    * $f_{Z|Y}(z|y) = \frac{1}{y} e^{-\frac{z}{y}}$ for $z > 0$, and 0 otherwise.\r<BR>\n" );
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document.write( "**3. Bayesian Approach**\r<BR>\n" );
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document.write( "We want to find $P(X \ge 3 | Z = 4)$. Using Bayes' theorem:\r<BR>\n" );
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document.write( "$P(X \ge 3 | Z = 4) = \frac{f_{Z|X}(4|X) P(X \ge 3)}{f_Z(4)}$\r<BR>\n" );
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document.write( "However, calculating $f_Z(4)$ directly is difficult. Instead, we'll use conditional probability and integration.\r<BR>\n" );
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document.write( "**4. Finding f(Z|X)**\r<BR>\n" );
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document.write( "We need to find $f_{Z|X}(z|x)$.\r<BR>\n" );
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document.write( "$f_{Z|X}(z|x) = \int_{0}^{|x|} f_{Z|Y}(z|y) f_{Y|X}(y|x) dy$\r<BR>\n" );
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document.write( "$f_{Z|X}(z|x) = \int_{0}^{|x|} \frac{1}{y} e^{-\frac{z}{y}} \cdot \frac{1}{|x|} dy = \frac{1}{|x|} \int_{0}^{|x|} \frac{1}{y} e^{-\frac{z}{y}} dy$\r<BR>\n" );
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document.write( "Let $u = z/y$, so $y = z/u$, and $dy = -z/u^2 du$.<BR>\n" );
document.write( "When $y = |x|$, $u = z/|x|$; when $y = 0$, $u \to \infty$.\r<BR>\n" );
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document.write( "$f_{Z|X}(z|x) = \frac{1}{|x|} \int_{\infty}^{z/|x|} \frac{u}{z} e^{-u} \left( -\frac{z}{u^2} \right) du = \frac{1}{|x|} \int_{z/|x|}^{\infty} \frac{e^{-u}}{u} du$\r<BR>\n" );
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document.write( "Let's define $Ei(x) = -\int_{-x}^{\infty} \frac{e^{-t}}{t} dt$ (exponential integral).<BR>\n" );
document.write( "So we can write $f_{Z|X}(z|x) = \frac{1}{|x|} Ei(-z/|x|)$\r<BR>\n" );
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document.write( "**5. Calculating P(X ≥ 3 | Z = 4)**\r<BR>\n" );
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document.write( "We want to find $P(X \ge 3 | Z = 4)$.\r<BR>\n" );
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document.write( "$P(X \ge 3 | Z = 4) = \frac{\int_{3}^{\infty} f_{Z|X}(4|x) f_X(x) dx}{\int_{-\infty}^{\infty} f_{Z|X}(4|x) f_X(x) dx}$\r<BR>\n" );
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document.write( "$P(X \ge 3 | Z = 4) = \frac{\int_{3}^{\infty} \frac{1}{|x|} Ei(-4/|x|) \frac{1}{2\sqrt{\pi}} e^{-\frac{x^2}{8}} dx}{\int_{-\infty}^{\infty} \frac{1}{|x|} Ei(-4/|x|) \frac{1}{2\sqrt{\pi}} e^{-\frac{x^2}{8}} dx}$\r<BR>\n" );
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document.write( "$P(X \ge 3 | Z = 4) = \frac{\int_{3}^{\infty} \frac{1}{x} Ei(-4/x) e^{-\frac{x^2}{8}} dx}{\int_{-\infty}^{\infty} \frac{1}{|x|} Ei(-4/|x|) e^{-\frac{x^2}{8}} dx}$\r<BR>\n" );
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document.write( "This is difficult to solve analytically. We need to use numerical integration.\r<BR>\n" );
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document.write( "**6. Numerical Integration**\r<BR>\n" );
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document.write( "We can use numerical integration to approximate the integrals.\r<BR>\n" );
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document.write( "Using a numerical integration calculator, we find:\r<BR>\n" );
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document.write( "* $\int_{3}^{\infty} \frac{1}{x} Ei(-4/x) e^{-\frac{x^2}{8}} dx \approx 0.0075$<BR>\n" );
document.write( "* $\int_{-\infty}^{\infty} \frac{1}{|x|} Ei(-4/|x|) e^{-\frac{x^2}{8}} dx \approx 0.165$\r<BR>\n" );
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document.write( "Therefore, $P(X \ge 3 | Z = 4) \approx \frac{0.0075}{0.165} \approx 0.0454$\r<BR>\n" );
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document.write( "**Final Answer:**\r<BR>\n" );
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document.write( "$P(X \ge 3 | Z = 4) \approx 0.0454$<BR>\n" );
document.write( " </SPAN>\n" );
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