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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

A person is given m dollars, which he must allocate between an event A and its complement\({{\bf{A}}^{\bf{c}}}\). Suppose that he allocates a dollars to A and m-a dollars to\({{\bf{A}}^{\bf{c}}}\). The person鈥檚 gain is then determined as follows: If A occurs, his gain is\({g_1}a\); if\({A^c}\)occurs, his gain is\({{\bf{g}}_{\bf{2}}}\left( {{\bf{m - a}}} \right)\)Here,\({{\bf{g}}_{\bf{1}}}\,{{\bf{g}}_{\bf{2}}}\)are given positive constants. Suppose also that\({\bf{{\rm P}}}\left( {\bf{A}} \right){\bf{ = p}}\)and the person鈥檚 utility function is\({\bf{U}}\left( {\bf{x}} \right){\bf{ = logx}}\)for x>0. Determine the amount a that will maximize the person鈥檚 expected utility, and show that this amount does not depend on the values of g1 and g2.

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b. Find the value at risk (VaR) for the example portfolio at probability level 0.97. Hint: Review Example 3.9.5 to see how to find the p.d.f. of the sum of two uniform random variables.

Suppose that\({\bf{X}}\),\({\bf{Y}}\),and\({\bf{Z}}\)are three random variables such that\({\bf{Var}}\left( {\bf{X}} \right){\bf{ = 1}}\),\({\bf{Var}}\left( {\bf{Y}} \right){\bf{ = 4}}\),\({\bf{Var}}\left( {\bf{Z}} \right){\bf{ = 8}}\),\({\bf{Cov}}\left( {{\bf{X,Y}}} \right){\bf{ = 1}}\),\({\bf{Cov}}\left( {{\bf{X,Z}}} \right){\bf{ = - 1}}\),and\({\bf{Cov}}\left( {{\bf{Y,Z}}} \right){\bf{ = 2}}\). Determine (a)\({\bf{Var}}\left( {{\bf{X + Y + Z}}} \right)\)and (b)\({\bf{Var}}\left( {{\bf{3X - Y - 2Z + 1}}} \right)\).

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\(\begin{align}{\bf{Pr}}\left( {{\bf{X = - 1}}} \right){\bf{ = 0}}{\bf{.6,}}\,{\bf{Pr}}\left( {{\bf{X = 0}}} \right){\bf{ = 0}}{\bf{.2,}}\,{\bf{Pr}}\left( {{\bf{X = 2}}} \right){\bf{ = 0}}{\bf{.2;}}\\{\bf{Pr}}\left( {{\bf{Y = 0}}} \right){\bf{ = 0}}{\bf{.9,}}\,\,\,\,{\bf{Pr}}\left( {{\bf{Y = 1}}} \right){\bf{ = 0}}{\bf{.1}}\end{align}\)

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