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Chapter 3: Q9E (page 187)

Suppose that the n variables\({{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{n}}}\) form a random sample from the uniform distribution on the interval [0, 1]and that the random variables \({{\bf{Y}}_{\bf{1}}}\;{\bf{and}}\;{{\bf{Y}}_{\bf{n}}}\) are defined as in Eq. (3.9.8). Determine the value of \({\bf{Pr}}\left( {{{\bf{Y}}_{\bf{1}}} \le {\bf{0}}{\bf{.1}}\;{\bf{and}}\;{\bf{Y}}_{\bf{n}}^{} \le {\bf{0}}{\bf{.8}}} \right)\)

Short Answer

Expert verified

\[{0.8^n} - {0.7^n}\]

Step by step solution

01

Given information

Here,\({X_1} \ldots {X_n}\)is a random sample from uniform distribution\(U\left( {0,1} \right)\).

\(\begin{aligned}{Y_1} &= \min \left\{ {{X_1} \ldots {X_n}} \right\}\\{Y_n} &= \max \left\{ {{X_1} \ldots {X_n}} \right\}\end{aligned}\)

02

Obtain the PDF and CDF of X

The pdf of a uniform distribution is obtained by using the formula: \(\frac{1}{{b - a}};a \le x \le b\).

Here, \(a = 0,b = 1\).

Therefore, the PDF of X is expressed as,

\({f_x} = \left\{ \begin{array}{l}\frac{1}{{1 - 0}} = 1\;\;\;\;\;\;\;\;\;\;0 \le x \le 1\\0;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{otherwise}}\end{array} \right.\)

The CDF of a uniform distribution is obtained by using the formula:

\(\begin{aligned}{F_X}\left( x \right) &= P\left( {X \le x} \right)\\ &= \frac{{x - a}}{{b - a}}\\ &= \frac{{x - 0}}{{1 - 0}}\\ &= x\end{aligned}\)

03

Defining the probability

We have to find,

\(\begin{aligned}\Pr \left( {{Y_1} \le 0.1\;{\rm{and}}\;Y_n^{} \le 0.8} \right) &= P\left( {Y_n^{} \le 0.8} \right) - P\left( {{Y_1} > 0.1,Y_n^{} \le 0.8} \right)\\ &= P\left( {{X_1} \le 0.8, \ldots ,{X_n} \le 0.8} \right) - P\left( {0.1 < {X_1} \le 0.8, \ldots ,0.1 < {X_n} \le 0.8} \right)\\ &= {\left[ {P\left( {{X_1} \le 0.8} \right)} \right]^n} - {\left[ {P\left( {0.1 < {X_1} \le 0.8} \right)} \right]^n}\\ &= {\left[ {F\left( {0.8} \right)} \right]^n} - {\left[ {F\left( {0.8} \right) - F\left( {0.1} \right)} \right]^n}\\ &= {0.8^n} - {\left( {0.8 - 0.1} \right)^n}\\ &= {0.8^n} - {0.7^n}\end{aligned}\)

Hence, the answer is\({\bf{0}}{\bf{.}}{{\bf{8}}^{\bf{n}}}{\bf{ - 0}}{\bf{.}}{{\bf{7}}^{\bf{n}}}\)

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Most popular questions from this chapter

Suppose that \({{\bf{X}}_{\bf{1}}}\;{\bf{and}}\;{{\bf{X}}_{\bf{2}}}\) are i.i.d. random variables andthat each of them has a uniform distribution on theinterval [0, 1]. Find the p.d.f. of\({\bf{Y = }}{{\bf{X}}_{\bf{1}}}{\bf{ + }}{{\bf{X}}_{\bf{2}}}\).

Suppose that a Markov chain has four states 1, 2, 3, 4, and stationary transition probabilities as specified by the following transition matrix

\(p = \left[ {\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{1}{4}}&0&{\frac{1}{2}}\\0&1&0&0\\{\frac{1}{2}}&0&{\frac{1}{2}}&0\\{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\end{array}} \right]\):

a.If the chain is in state 3 at a given timen, what is the probability that it will be in state 2 at timen+2?

b.If the chain is in state 1 at a given timen, what is the probability it will be in state 3 at timen+3?

Suppose that a random variableXhas the binomial distribution with parametersn=15 andp=0.5. Find Pr(X <6).

Question:Suppose that the joint p.d.f. ofXandYis as follows:

\(\)\({\bf{f}}\left( {{\bf{x,y}}} \right){\bf{ = }}\left\{ \begin{array}{l}\frac{{{\bf{15}}}}{{\bf{4}}}{{\bf{x}}^{\bf{2}}}\;{\bf{for}}\;{\bf{0}} \le {\bf{y}} \le {\bf{1 - }}{{\bf{x}}^{\bf{2}}}\\{\bf{0}}\;{\bf{otherwise}}\end{array} \right.\)

a. Determine the marginal p.d.f.鈥檚 ofXandY.

b. AreXandYindependent?

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\({{\bf{y}}_{\bf{1}}}\).
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