document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1164407>Question 1164407</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The random variables (X,Y) have joint pdf f(x,y) = 1/x for the region 0 < x < 1, 0<y<x. The pdf is zero outside this region. <BR>\n" );
document.write( "iii. Find the covariance of X, Y i.e. cov(X,Y)<BR>\n" );
document.write( "iv. Find P(X^2+Y^2 ≤ 1| X=x)<BR>\n" );
document.write( "v. Find P(X^2+Y^2 ≤ 1) <BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #854350 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2264)</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=854350\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To find the covariance and the probabilities for the joint density function $f(x,y) = \frac{1}{x}$ on the region $0 < x < 1, 0 < y < x$, we proceed with the following evaluations.\r<BR>\n" );
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document.write( "### **iii. Find the Covariance of $X$ and $Y$**\r<BR>\n" );
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document.write( "The covariance is defined as:<BR>\n" );
document.write( "$$\text{Cov}(X, Y) = E[XY] - E[X]E[Y]$$\r<BR>\n" );
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document.write( "**1. Calculate $E[X]$:**<BR>\n" );
document.write( "We first find the marginal density $f_X(x)$:<BR>\n" );
document.write( "$$f_X(x) = \int_0^x \frac{1}{x} \, dy = \left[ \frac{y}{x} \right]_0^x = 1, \quad 0 < x < 1$$<BR>\n" );
document.write( "Now, $E[X] = \int_0^1 x \cdot f_X(x) \, dx = \int_0^1 x \cdot 1 \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}$\r<BR>\n" );
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document.write( "**2. Calculate $E[Y]$:**<BR>\n" );
document.write( "$$E[Y] = \int_0^1 \int_0^x y \cdot \frac{1}{x} \, dy \, dx = \int_0^1 \left[ \frac{y^2}{2x} \right]_0^x \, dx = \int_0^1 \frac{x^2}{2x} \, dx = \int_0^1 \frac{x}{2} \, dx = \left[ \frac{x^2}{4} \right]_0^1 = \frac{1}{4}$$\r<BR>\n" );
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document.write( "**3. Calculate $E[XY]$:**<BR>\n" );
document.write( "$$E[XY] = \int_0^1 \int_0^x xy \cdot \frac{1}{x} \, dy \, dx = \int_0^1 \int_0^x y \, dy \, dx = \int_0^1 \left[ \frac{y^2}{2} \right]_0^x \, dx = \int_0^1 \frac{x^2}{2} \, dx = \left[ \frac{x^3}{6} \right]_0^1 = \frac{1}{6}$$\r<BR>\n" );
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document.write( "**4. Compute Covariance:**<BR>\n" );
document.write( "$$\text{Cov}(X, Y) = \frac{1}{6} - \left( \frac{1}{2} \cdot \frac{1}{4} \right) = \frac{1}{6} - \frac{1}{8} = \frac{4 - 3}{24} = \frac{1}{24}$$\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### **iv. Find $P(X^2 + Y^2 \le 1 \mid X = x)$**\r<BR>\n" );
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document.write( "Given $X = x$, we look at the conditional density $f(y|x) = \frac{f(x,y)}{f_X(x)} = \frac{1/x}{1} = \frac{1}{x}$ for $0 < y < x$.<BR>\n" );
document.write( "The condition $x^2 + y^2 \le 1$ becomes $y^2 \le 1 - x^2$, or **$y \le \sqrt{1 - x^2}$**.\r<BR>\n" );
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document.write( "* If $x \le \frac{1}{\sqrt{2}}$, then $x \le \sqrt{1 - x^2}$. The entire range of $y$ (from $0$ to $x$) satisfies the condition. Thus, **$P = 1$**.<BR>\n" );
document.write( "* If $x > \frac{1}{\sqrt{2}}$, the condition $y \le \sqrt{1 - x^2}$ is more restrictive than $y < x$. <BR>\n" );
document.write( "    $$P = \int_0^{\sqrt{1 - x^2}} \frac{1}{x} \, dy = \frac{\sqrt{1 - x^2}}{x}$$\r<BR>\n" );
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document.write( "**Result:**<BR>\n" );
document.write( "$$P(X^2 + Y^2 \le 1 \mid X = x) = <BR>\n" );
document.write( "\begin{cases} <BR>\n" );
document.write( "1 & 0 < x \le \frac{1}{\sqrt{2}} \\<BR>\n" );
document.write( "\frac{\sqrt{1 - x^2}}{x} & \frac{1}{\sqrt{2}} < x < 1<BR>\n" );
document.write( "\end{cases}$$\r<BR>\n" );
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document.write( "### **v. Find $P(X^2 + Y^2 \le 1)$**\r<BR>\n" );
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document.write( "This is the integral of the conditional probability over the marginal density of $X$:<BR>\n" );
document.write( "$$P = \int_0^{1/\sqrt{2}} (1)(1) \, dx + \int_{1/\sqrt{2}}^1 \frac{\sqrt{1 - x^2}}{x} (1) \, dx$$\r<BR>\n" );
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document.write( "**Calculating the integrals:**<BR>\n" );
document.write( "1.  **Term 1:** $\int_0^{1/\sqrt{2}} 1 \, dx = \frac{1}{\sqrt{2}}$<BR>\n" );
document.write( "2.  **Term 2:** Using the substitution $x = \sin\theta$:<BR>\n" );
document.write( "    $$\int \frac{\cos\theta}{\sin\theta} \cos\theta \, d\theta = \ln| \csc\theta - \cot\theta | + \cos\theta$$<BR>\n" );
document.write( "    Evaluating from $x = 1/\sqrt{2}$ ($\theta = \pi/4$) to $x = 1$ ($\theta = \pi/2$):<BR>\n" );
document.write( "    $$[0] - [\ln(\sqrt{2}-1) + 1/\sqrt{2}] = -\ln(\sqrt{2}-1) - \frac{1}{\sqrt{2}}$$\r<BR>\n" );
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document.write( "**Total Probability:**<BR>\n" );
document.write( "$$P = \frac{1}{\sqrt{2}} - \ln(\sqrt{2}-1) - \frac{1}{\sqrt{2}} = -\ln(\sqrt{2}-1) = \ln\left(\frac{1}{\sqrt{2}-1}\right) = \ln(1+\sqrt{2})$$<BR>\n" );
document.write( "$$P \approx 0.8814$$ </SPAN>\n" );
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