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Chapter 3: Q 11E (page 107)

Show that there does not exist any numbercsuch that the following functionf (x)would be a p.d.f.:

Short Answer

Expert verified

There does not exist any number c such that


is a probability density function.

Step by step solution

01

Given the information

The function is given as,


02

Calculate the value of c 

For a function to be a probability density function, the function should hold the following condition.

\int\limits_x^{}{f\left(x\right)}=1

For the provided function, the calculations are as follows:

Since, In (鈭� ) = 鈭� and In (0) is not defined, there does not exist any number c such that the provided function is a probability density function.

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Most popular questions from this chapter

Two students,AandB,are both registered for a certain course. Assume that studentAattends class 80 percent of the time, studentBattends class 60 percent of the time, and the absences of the two students are independent. Consider the conditions of Exercise 7 of Sec. 2.2 again. If exactly one of the two students,AandB,is in class on a given day, what is the probability that it isA?

Suppose thatnletters are placed at random innenvelopes, as in the matching problem of Sec. 1.10, and letqndenote the probability that no letter is placed in the correct envelope. Consider the conditions of Exercise 30 again. Show that the probability that exactly two letters are placed in

the correct envelopes are(1/2) qn鈭�2

Show that there does not exist any numbercsuch that the following function would be a p.f.:

\(f\left( x \right) = \left\{ \begin{array}{l}\frac{c}{x}\;\;\;\;for\;x = 1,2,...\\0\;\;\;\;otherwise\end{array} \right.\)

The unique stationary distribution in Exercise 9 is \({\bf{v = }}\left( {{\bf{0,1,0,0}}} \right)\). This is an instance of the following general result: Suppose that a Markov chain has exactly one absorbing state. Suppose further that, for each non-absorbing state \({\bf{k}}\), there is \({\bf{n}}\) such that the probability is positive of moving from state \({\bf{k}}\) to the absorbing state in \({\bf{n}}\) steps. Then the unique stationary distribution has probability 1 in the absorbing state. Prove this result.

Question:Suppose that in a certain drug the concentration of aparticular chemical is a random variable with a continuousdistribution for which the p.d.f.gis as follows:

\({\bf{g}}\left( {\bf{x}} \right){\bf{ = }}\left\{ \begin{array}{l}\frac{{\bf{3}}}{{\bf{8}}}{{\bf{x}}^{\bf{2}}}\;{\bf{for}}\;{\bf{0}} \le {\bf{x}} \le {\bf{2}}\\{\bf{0}}\;{\bf{otherwise}}\end{array} \right.\)

Suppose that the concentrationsXandYof the chemicalin two separate batches of the drug are independent randomvariables for each of which the p.d.f. isg. Determine

(a) the joint p.d.f.of X andY;

(b) Pr(X=Y);

(c) Pr(X >Y );

(d) Pr(X+Y鈮�1).
See all solutions

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