document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1164488>Question 1164488</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( "iv. Find P(X^2+Y^2 ≤ 1| X=x)<BR>\n" );
document.write( "v. Find P(X^2+Y^2 ≤ 1) <BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #854347 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=854347\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> ```python?code_reference&code_event_index=1<BR>\n" );
document.write( "import numpy as np\r<BR>\n" );
document.write( "\n" );
document.write( "# Region: 0 < x < 1, 0 < y < x<BR>\n" );
document.write( "# f(x, y) = 1/x\r<BR>\n" );
document.write( "\n" );
document.write( "# Verify total probability<BR>\n" );
document.write( "# integral_{0}^{1} integral_{0}^{x} (1/x) dy dx<BR>\n" );
document.write( "# = integral_{0}^{1} [y/x]_{0}^{x} dx<BR>\n" );
document.write( "# = integral_{0}^{1} 1 dx = 1. Correct.\r<BR>\n" );
document.write( "\n" );
document.write( "# iv. P(X^2 + Y^2 <= 1 | X=x)<BR>\n" );
document.write( "# Conditional pdf f(y|x) = f(x,y) / f_X(x)<BR>\n" );
document.write( "# f_X(x) = integral_{0}^{x} (1/x) dy = 1 for 0 < x < 1.<BR>\n" );
document.write( "# So f(y|x) = (1/x) / 1 = 1/x for 0 < y < x.<BR>\n" );
document.write( "# We want P(Y^2 <= 1 - x^2 | X=x) = P(Y <= sqrt(1 - x^2) | X=x)<BR>\n" );
document.write( "# Since the region is 0 < y < x:<BR>\n" );
document.write( "# If sqrt(1 - x^2) >= x, then the entire range 0 < y < x is included.<BR>\n" );
document.write( "# sqrt(1 - x^2) >= x  => 1 - x^2 >= x^2 => 1 >= 2x^2 => x <= 1/sqrt(2)<BR>\n" );
document.write( "# If x <= 1/sqrt(2), P = 1.<BR>\n" );
document.write( "# If x > 1/sqrt(2), P = integral_{0}^{sqrt(1-x^2)} (1/x) dy = sqrt(1-x^2)/x.\r<BR>\n" );
document.write( "\n" );
document.write( "# v. P(X^2 + Y^2 <= 1)<BR>\n" );
document.write( "# Total probability.<BR>\n" );
document.write( "# P = integral_{0}^{1} P(X^2+Y^2 <= 1 | X=x) * f_X(x) dx<BR>\n" );
document.write( "# P = integral_{0}^{1/sqrt(2)} 1 dx + integral_{1/sqrt(2)}^{1} (sqrt(1-x^2)/x) dx\r<BR>\n" );
document.write( "\n" );
document.write( "from scipy.integrate import quad\r<BR>\n" );
document.write( "\n" );
document.write( "term1 = 1/np.sqrt(2)<BR>\n" );
document.write( "term2, _ = quad(lambda x: np.sqrt(1 - x**2) / x, 1/np.sqrt(2), 1)\r<BR>\n" );
document.write( "\n" );
document.write( "total_prob = term1 + term2<BR>\n" );
document.write( "print(f\"Term 1: {term1}\")<BR>\n" );
document.write( "print(f\"Term 2: {term2}\")<BR>\n" );
document.write( "print(f\"Total Probability: {total_prob}\")\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "```<BR>\n" );
document.write( "```text?code_stdout&code_event_index=1<BR>\n" );
document.write( "Term 1: 0.7071067811865475<BR>\n" );
document.write( "Term 2: 0.1742668058329955<BR>\n" );
document.write( "Total Probability: 0.8813735870195429\r<BR>\n" );
document.write( "\n" );
document.write( "```<BR>\n" );
document.write( "To solve these problems, we first need the marginal density $f_X(x)$ and the conditional density $f(y|x)$.\r<BR>\n" );
document.write( "\n" );
document.write( "**Given:**<BR>\n" );
document.write( "* $f(x, y) = \frac{1}{x}$ for $0 < x < 1$ and $0 < y < x$.<BR>\n" );
document.write( "* The marginal density of $X$: $f_X(x) = \int_0^x \frac{1}{x} \, dy = [\frac{y}{x}]_0^x = 1$ for $0 < x < 1$.<BR>\n" );
document.write( "* The conditional density of $Y$ given $X=x$: $f(y|x) = \frac{f(x,y)}{f_X(x)} = \frac{1/x}{1} = \frac{1}{x}$ for $0 < y < x$.\r<BR>\n" );
document.write( "\n" );
document.write( "---\r<BR>\n" );
document.write( "\n" );
document.write( "### **iv. Find $P(X^2 + Y^2 \le 1 \mid X = x)$**\r<BR>\n" );
document.write( "\n" );
document.write( "This is the probability that $Y^2 \le 1 - x^2$ given a fixed $x$. This simplifies to $Y \le \sqrt{1 - x^2}$.<BR>\n" );
document.write( "Since our variable $Y$ is bounded by $0 < y < x$, we have two cases based on whether the circle boundary $\sqrt{1 - x^2}$ is inside or outside the range of $y$:\r<BR>\n" );
document.write( "\n" );
document.write( "1.  **Case 1: $x \le \sqrt{1 - x^2}$**<BR>\n" );
document.write( "    This occurs when $x^2 \le 1 - x^2 \implies 2x^2 \le 1 \implies x \le \frac{1}{\sqrt{2}}$.<BR>\n" );
document.write( "    In this case, the entire range of $y$ ($0 < y < x$) satisfies the condition.<BR>\n" );
document.write( "    $$P(X^2 + Y^2 \le 1 \mid X = x) = 1$$\r<BR>\n" );
document.write( "\n" );
document.write( "2.  **Case 2: $x > \frac{1}{\sqrt{2}}$**<BR>\n" );
document.write( "    The condition $Y \le \sqrt{1 - x^2}$ is more restrictive than the boundary $y < x$.<BR>\n" );
document.write( "    $$P(X^2 + Y^2 \le 1 \mid X = x) = \int_0^{\sqrt{1-x^2}} \frac{1}{x} \, dy = \frac{\sqrt{1-x^2}}{x}$$\r<BR>\n" );
document.write( "\n" );
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" );
document.write( "\n" );
document.write( "---\r<BR>\n" );
document.write( "\n" );
document.write( "### **v. Find $P(X^2 + Y^2 \le 1)$**\r<BR>\n" );
document.write( "\n" );
document.write( "We calculate this by integrating the conditional probability over the marginal density of $x$:<BR>\n" );
document.write( "$$P(X^2 + Y^2 \le 1) = \int_0^1 P(X^2 + Y^2 \le 1 \mid X = x) f_X(x) \, dx$$<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" );
document.write( "\n" );
document.write( "**Solving the integrals:**<BR>\n" );
document.write( "1.  **First Part:** $\int_0^{1/\sqrt{2}} 1 \, dx = \frac{1}{\sqrt{2}} \approx 0.707$<BR>\n" );
document.write( "2.  **Second Part:** For $\int \frac{\sqrt{1-x^2}}{x} dx$, we use the substitution $x = \sin\theta$:<BR>\n" );
document.write( "    $$\int \frac{\cos\theta}{\sin\theta} \cos\theta \, d\theta = \int (\csc\theta - \sin\theta) \, d\theta = \ln|\csc\theta - \cot\theta| + \cos\theta$$<BR>\n" );
document.write( "    Evaluating this from $x = \frac{1}{\sqrt{2}}$ ($\theta = \frac{\pi}{4}$) to $x = 1$ ($\theta = \frac{\pi}{2}$):<BR>\n" );
document.write( "    $$\left[ \ln\left| \frac{1 - \sqrt{1-x^2}}{x} \right| + \sqrt{1-x^2} \right]_{1/\sqrt{2}}^1 = (0 + 0) - (\ln(\sqrt{2}-1) + \frac{1}{\sqrt{2}})$$<BR>\n" );
document.write( "    $$\text{Second Part} = -\ln(\sqrt{2}-1) - \frac{1}{\sqrt{2}} \approx 0.881 - 0.707 = 0.174$$\r<BR>\n" );
document.write( "\n" );
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( "$$P \approx 0.8814$$\r<BR>\n" );
document.write( "\n" );
document.write( "**Result:**<BR>\n" );
document.write( "$$P(X^2 + Y^2 \le 1) = \ln(1 + \sqrt{2}) \approx 0.8814$$ </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );