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Suppose thatXhas the p.d.f.


Find and sketch the c.d.f. ofX.

Short Answer

Expert verified

The c.d.f of X is

Step by step solution

01

Given information


02

Calculate c.d.f of X

03

Drawing graph of c.d.f

The c.d.f of X is given by:


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Most popular questions from this chapter

Suppose 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(Xt)=1/4.

b. Find the value oftsuch that Pr(Xt)=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?

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