Chapter 3: Q15E (page 117)
Suppose thatXhas the p.d.f.
Find and sketch the c.d.f. ofX.
Short Answer
The c.d.f of X is
Chapter 3: Q15E (page 117)
Suppose thatXhas the p.d.f.
Find and sketch the c.d.f. ofX.
The c.d.f of X is
All the tools & learning materials you need for study success - in one app.
Get started for freeSuppose that the p.d.f. of a random variableXis as follows:
\(f\left( x \right) = \left\{ \begin{array}{l}\frac{1}{8}x\;\;for\;0 \le x \le 4\\0\;\;\;\;otherwise\end{array} \right.\)
a. Find the value oftsuch that Pr(X≤t)=1/4.
b. Find the value oftsuch that Pr(X≥t)=1/2.
Consider the situation described in Example 3.7.14. Suppose that \(\) \({X_1} = 5\) and\({X_2} = 7\)are observed.
a. Compute the conditional p.d.f. of \({X_3}\) given \(\left( {{X_1},{X_2}} \right) = \left( {5,7} \right)\).
b. Find the conditional probability that \({X_3} > 3\)given \(\left( {{X_1},{X_2}} \right) = \left( {5,7} \right)\)and compare it to the value of \(P\left( {{X_3} > 3} \right)\)found in Example 3.7.9. Can you suggest a reason why the conditional probability should be higher than the marginal probability?
For the conditions of Exercise 1, find the p.d.f. of the
average \(\frac{{\left( {{{\bf{X}}_{\bf{1}}}{\bf{ + }}{{\bf{X}}_{\bf{2}}}} \right)}}{{\bf{2}}}\)
The definition of the conditional p.d.f. of X given\({\bf{Y = y}}\)is arbitrary if\({{\bf{f}}_{\bf{2}}}\left( {\bf{y}} \right){\bf{ = 0}}\). The reason that this causes no serious problem is that it is highly unlikely that we will observe Y close to a value\({{\bf{y}}_{\bf{0}}}\)such that\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{0}}}} \right){\bf{ = 0}}\). To be more precise, let\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{0}}}} \right){\bf{ = 0}}\), and let\({{\bf{A}}_{\bf{0}}}{\bf{ = }}\left( {{{\bf{y}}_{\bf{0}}}{\bf{ - }} \in {\bf{,}}{{\bf{y}}_{\bf{0}}}{\bf{ + }} \in } \right)\). Also, let\({{\bf{y}}_{\bf{1}}}\)be such that\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{1}}}} \right){\bf{ > 0}}\), and let\({{\bf{A}}_{\bf{1}}}{\bf{ = }}\left( {{{\bf{y}}_{\bf{1}}}{\bf{ - }} \in {\bf{,}}{{\bf{y}}_{\bf{1}}}{\bf{ + }} \in } \right)\). Assume that\({{\bf{f}}_{\bf{2}}}\)is continuous at both\({{\bf{y}}_{\bf{0}}}\)and\({{\bf{y}}_{\bf{1}}}\).
Show that
\(\mathop {{\bf{lim}}}\limits_{ \in \to {\bf{0}}} \,\frac{{{\bf{Pr}}\left( {{\bf{Y}} \in {{\bf{A}}_{\bf{0}}}} \right)}}{{{\bf{Pr}}\left( {{\bf{Y}} \in {{\bf{A}}_{\bf{1}}}} \right)}}{\bf{ = 0}}{\bf{.}}\)
That is, the probability that Y is close to\({{\bf{y}}_{\bf{0}}}\)is much smaller than the probability that Y is close to\({{\bf{y}}_{\bf{1}}}\).
Question:Suppose that the joint p.d.f. ofXandYis as follows:
\(f\left( {x,y} \right) = \left\{ \begin{array}{l}2x{e^{ - y}}\;for\;0 \le x \le 1\;and\;0 < y < \infty \\0\;otherwise\end{array} \right.\)
AreXandYindependent?
What do you think about this solution?
We value your feedback to improve our textbook solutions.