/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q25SE In this exercise, we shall provi... [FREE SOLUTION] | 91Ó°ÊÓ

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In this exercise, we shall provide an approximate justification for Eq. (3.6.6). First, remember that if a and bare close together, then

\(\int\limits_a^b {r\left( t \right)} dt \approx \left( {b - a} \right)r\left( {\frac{{a + b}}{2}} \right)\)

Throughout this problem, assume that X and Y have joint pdf f .

a. Use (3.11.1) to approximate\({\rm P}\left( {y - \varepsilon < Y \le y + \varepsilon } \right)\)

b. Use (3.11.1) with\(r\left( t \right) = f\left( {s,t} \right)\)for fixed s to approximate

\({\rm P}\left( {X \le x\,and\,y - \varepsilon < Y \le y + \varepsilon } \right)\)

\( = \int\limits_{ - \infty }^x {\int\limits_{y - \varepsilon }^{y + \varepsilon } {f\left( {s,t} \right)dtds} } \).

c. Show that the ratio of the approximation in part (b) to the approximation in part (a) is\(\int\limits_{ - \infty }^x {{g_1}\left( {s|y} \right)ds.} \)

Short Answer

Expert verified
  1. \({\rm P}\left( {y - \varepsilon < Y \le y + \varepsilon } \right)\;\; \approx 2\varepsilon {f_2}\left( y \right)\)
  2. \({\rm P}\left( {X \le x\,and\,y - \varepsilon < Y \le y + \varepsilon } \right) \approx 2\varepsilon \int\limits_{ - \infty }^\infty {f\left( {s,y} \right)ds} \)
  3. The ratio of the approximation of part (b) to (a) is \(\int\limits_{ - \infty }^x {{g_1}\left( {s|y} \right)ds.} \)

Step by step solution

01

Given information

A and b are close together then

\(\int\limits_a^b {r\left( t \right)} dt \approx \left( {b - a} \right)r\left( {\frac{{a + b}}{2}} \right)\)

02

Approximating the probabilities.

X and y have the joint pdf f

The conditional pdf of\({g_1}\left( {x,y} \right)\):

\({g_1}\left( {x,y} \right) = \mathop {\lim }\limits_{\varepsilon \to 0} \frac{\partial }{{\partial x}}{\rm P}\left( {X \le x|y - \varepsilon < Y \le y + \varepsilon } \right)\)

\({f_2}\)be the marginal pdf of Y.

Hence we approximate

\({\rm P}\left( {y - \varepsilon < Y \le y + \varepsilon } \right) = \int\limits_{y - \varepsilon }^{y + \varepsilon } {{f_2}\left( t \right)dt} .\)

\( \approx 2\varepsilon {f_2}\left( y \right)\)

Step 3: When\({\rm P}\left( {X \le x\,and\,y - \varepsilon < Y \le y + \varepsilon } \right)\)and\( = \int\limits_{ - \infty }^x {\int\limits_{y - \varepsilon }^{y + \varepsilon } {f\left( {s,t} \right)dtds} } \)

For each s, we approximate

\(\int\limits_{y - \varepsilon }^{y + \varepsilon } {f\left( {s,t} \right)dt \approx 2\varepsilon f\left( {s,y} \right)} .\)

\(\int\limits_a^b {r\left( t \right)} dt \approx \left( {b - a} \right)r\left( {\frac{{a + b}}{2}} \right)\)with\(r\left( t \right) = f\left( {s,t} \right)\)

Therefore

\({\rm P}\left( {X \le x\,and\,y - \varepsilon < Y \le y + \varepsilon } \right) = \int\limits_{ - \infty }^x {\int\limits_{y - \varepsilon }^{y + \varepsilon } {f\left( {s,t} \right)dtds} } \)

\( \approx 2\varepsilon \int\limits_{ - \infty }^\infty {f\left( {s,y} \right)ds} \)

03

To prove part (b)

Taking the ratio of the approximation in part(b) to the approximation in part (a)

We get,

\(\)\(\begin{aligned}{}{\rm P}\left( {X \le {\raise0.7ex\hbox{$x$} \!\mathord{\left/{\vphantom {x {y - \varepsilon}}}\right.}\!\lower0.7ex\hbox{${y - \varepsilon }$}}\, < Y \le y + \varepsilon } \right) &= \frac{{{\rm P}\left( {X \le x,y - \varepsilon \, < Y \le y + \varepsilon } \right)}}{{{\rm P}\left( {y - \varepsilon \, < Y \le \varepsilon } \right)}}\\ \approx \frac{{\int\limits_{ - \infty }^x {f\left( {s,y} \right)ds} }}{{{f_2}\left( y \right)}}\end{aligned}\)

So, \(P = \int\limits_{ - \infty }^x {{g_1}\left( {s|y} \right)} ds.\)

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