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Chapter 4: Q 1SE (page 272)

Suppose that the random variable X has a continuous distribution with c.d.f.\(F\left( x \right)\)and p.d.f. f. Suppose also that\(E\left( x \right)\)exists. Prove that\(\mathop {\lim }\limits_{x \to \infty } x\left( {1 - F\left( x \right)} \right) = 0\)

Hint: Use the fact that if E(X) exists, then

\(E\left( x \right) = \mathop {\lim }\limits_{\mu \to \infty } \int\limits_{ - \infty }^u {xf\left( x \right)} dx\)

Short Answer

Expert verified

\(\mathop {\lim }\limits_{x \to \infty } x\left( {1 - F\left( x \right)} \right) = 0\)

Step by step solution

01

Given information

X be the continuous distribution with cdf \(F\left( x \right)\)

02

verifying \(\mathop {\lim }\limits_{x \to \infty } x\left( {1 - F\left( x \right)} \right) = 0\)

Let \(E\left( x \right) = \mathop {\lim }\limits_{\mu \to \infty } \int\limits_{ - \infty }^\mu {xf\left( x \right)} dx\)

If\(u \ge 0\)

\(\begin{align}\int\limits_u^\infty {xf\left( x \right)dx} \ge u\int\limits_u^\infty {f\left( x \right)dx} \\ = u\left( {1 - F\left( u \right)} \right)\end{align}\)

Since

\(\begin{align}\mathop {\lim }\limits_{u \to \infty } \int\limits_{ - \infty }^u {xf\left( x \right)dx} &= E\left( x \right)\\ &= \int\limits_{ - \infty }^\infty {xf\left( x \right)dx} < \infty \end{align}\)

Hence, it follows that

\(\mathop {\lim }\limits_{u \to \infty } \left( {E\left( x \right) - \int\limits_{ - \infty }^u {xf\left( x \right)dx} } \right) = \)

Therefore,

\(\mathop {\lim }\limits_{u \to \infty } \int\limits_u^\infty {xf\left( x \right)dx} = 0\)

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Most popular questions from this chapter

Suppose that one letter is to be selected at random from the 30 letters in the sentence given in Exercise 4. If Y denotes the number of letters in the word in which the selected letter appears, what is the value of E (Y)?

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Hint: Youmay wish to use the formula \(\sum\limits_{k = 1}^n {{k^2}} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\)

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