Chapter 1: Problem 15
If \(E X_{1}^{-}<\infty\) and \(X_{n} \uparrow X\) then \(E X_{n} \uparrow E X\)
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Chapter 1: Problem 15
If \(E X_{1}^{-}<\infty\) and \(X_{n} \uparrow X\) then \(E X_{n} \uparrow E X\)
These are the key concepts you need to understand to accurately answer the question.
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Apply Jensen's inequality with \(\varphi(x)=e^{x}\) and \(P\left(X=\log y_{m}\right)=p(m)\) to conclude that if \(\sum_{m=1}^{n} p(m)=1\) and \(p(m), y_{m}>0\) then $$ \sum_{m=1}^{n} p(m) y_{m} \geq \prod_{m=1}^{n} y_{m}^{p(m)} $$ When \(p(m)=1 / n\), this says the arithmetic mean exceeds the geometric mean.
(i) Show that if \(X\) and \(Y\) are independent then \(\sigma(X)\) and \(\sigma(Y)\) are. (ii) Conversely, if \(\mathcal{F}\) and \(\mathcal{G}\) are independent, \(X \in \mathcal{F}\), and \(Y \in \mathcal{G}\), then \(X\) and \(Y\) are independent.
A useful lower bound. Let \(Y \geq 0\) with \(E Y^{2}<\infty .\) Apply the Cauchy- Schwarz inequality to \(Y 1_{(Y>0)}\) and conclude $$ P(Y>0) \geq(E Y)^{2} / E Y^{2} $$
Investment problem. We assume that at the beginning of each year you can buy bonds for \(\$ 1\) that are worth \(\$ a\) at the end of the year or stocks that are worth a random amount \(V \geq 0\). If you always invest a fixed proportion \(p\) of your wealth in bonds, then your wealth at the end of year \(n+1\) is \(W_{n+1}=(a p+\) \(\left.(1-p) V_{n}\right) W_{n}\). Suppose \(V_{1}, V_{2}, \ldots\) are i.i.d. with \(E V_{n}^{2}<\infty\) and \(E\left(V_{n}^{-2}\right)<\infty\) (i) Show that \(n^{-1} \log W_{n} \rightarrow c(p)\) a.s. (ii) Show that \(c(p)\) is concave. [Use (9.1) in the Appendix to justify differentiating under the expected value.] (iii) By investigating \(c^{\prime}(0)\) and \(c^{\prime}(1)\), give conditions on \(V\) that guarantee that the optimal choice of \(p\) is in \((0,1)\). (iv) Suppose \(P(V=1)=P(V=4)=1 / 2\). Find the optimal \(p\) as a function of \(a\).
Show that if \(F(x)=P(X \leq x)\) is continuous then \(Y=F(X)\) has a uniform distribution on \((0,1)\), that is, if \(y \in[0,1], P(Y \leq y)=y\).
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