Question

Let \(Y_{1}, Y_{2}, \ldots\) be independent identically distributed positive rando?n variables having finite mean \(\mu\). For fixed \(0<\beta<1\), let \(a\) be the smallest value \(u\) for which \(u \geq \beta E\left[u \vee Y_{1}\right]=\beta E\left[\max \left\\{u, Y_{1}\right\\}\right]\). Set \(f(x)=a \vee x\). Show that \(\left\\{\beta^{\prime \prime} f\left(M_{n}\right)\right\\}\) is a nonegative supermartingale, where \(M_{n}=\max \left\\{Y_{1}, \ldots, Y_{n}\right\\}\) whence \(a=f(0) \geq E\left[\beta^{T} f\left(M_{T}\right)\right]\) for all Markov times \(T\). Finally establish that \(a=E\left[\beta^{T} M_{T *}\right]\) for \(T^{*}=\min \left\\{n \geq 1: Y_{n} \geq a\right\\} .\) Thus, \(T^{*}\) maximizes \(E\left[\beta^{T} M_{T} \mid\right.\) over all Markov times \(T\).

Short answer

In summary, we have shown that \(\left\{\beta^n f(M_n)\right\}\) is a non-negative supermartingale and established that \(a \geq E\left[\beta^T f(M_T)\right]\) for all Markov times \(T\). Lastly, we have proved that \(a = E\left[\beta^{T^*} M_{T^*}\right]\) specifically for the Markov time \(T^*\), which maximizes \(E\left[\beta^T M_T\right]\) over all Markov times \(T\).

Step-by-step solution

1. Define Functions and Concepts

Define the function \(f(x) = a \vee x\), where \(a\) is the minimum value \(u\) such that \(u \geq \beta E[u \vee Y_1] = \beta E[\max\{u, Y_1\}]\). Define \(M_n = \max\{Y_1, \ldots, Y_n\}\) as the maximum value of the first \(n\) observations in the sequence. A non-negative supermartingale is a stochastic process \(\{X_n\}\) such that \(E[X_{n+1}\mid X_1, X_2, \ldots, X_n] \leq X_n\) and \(X_n \geq 0\) for all non-negative integer \(n\). Define the Markov time \(T^*\) as the first time when the random variable \(Y_n\) reaches or exceeds \(a\): \(T^* = \min\{n \geq 1: Y_n \geq a\}\).

2. Prove that \(\left\{\beta^n f(M_n)\right\}\) is a Non-negative Supermartingale

To show that \(\left\{\beta^n f(M_n)\right\}\) is a non-negative supermartingale, we must prove that: 1. \(E[\beta^{n+1}f(M_{n+1})\mid Y_1, \ldots Y_n] \leq \beta^n f(M_n)\) 2. \(\beta^n f(M_n) \geq 0\) Now, we have: \(E[\beta^{n+1}f(M_{n+1})\mid Y_1, \ldots Y_n] = \beta^{n+1}E[f(M_n \vee Y_{n+1})\mid Y_1, \ldots Y_n]\) Since \(Y_i\) are i.i.d., we have \(E[f(M_n \vee Y_{n+1})\mid Y_1, \ldots Y_n] = E[f(M_n \vee Y_1)]\). Since \(\beta a \leq \beta f(Y_1)\), by definition of \(a\): \(E[\beta^{n+1} f(M_{n+1})\mid Y_1, \ldots Y_n] \leq \beta^{n+1} E[f(M_n \vee Y_1)] \leq \beta^n f(M_n)\) Since \(f(x) = a \vee x\) and we know that \(a\) and \(x\) are non-negative, it follows that \(f(x) \geq 0\) for all \(x\). Thus, \(\beta^n f(M_n) \geq 0\). This proves that \(\left\{\beta^n f(M_n)\right\}\) is a non-negative supermartingale.

3. Show that \(a \geq E\left[\beta^T f(M_T)\right]\) for all Markov Times \(T\)

Following the properties of non-negative supermartingales: \(E[\beta^T f(M_T)] \leq E\left[\beta^0 f(M_0)\right] = E[f(0)] = a\) This proves that \(a \geq E\left[\beta^T f(M_T)\right]\) for all Markov times \(T\).

4. Establish that \(a = E\left[\beta^{T^*} M_{T^*}\right]\) for the Particular Time \(T^*\)

By definition, \(T^* = \min\{n \geq 1: Y_n \geq a\}\). We know that \(a \geq E\left[\beta^T f(M_T)\right]\) for all Markov times \(T\). When \(T=T^*\), \(f(M_{T^*}) = M_{T^*}\), since \(f(x) = a \vee x\). So, \(a \geq E\left[\beta^{T^*} M_{T^*}\right]\). However, since \(M_{T^*} \geq a\) for all \(T^*\): \(E\left[\beta^{T^*} M_{T^*}\right] \geq E\left[\beta^{T^*} a\right] = a E[\beta^{T^*}]\) Using both inequalities, we get: \(a \geq E\left[\beta^{T^*} M_{T^*}\right] \geq a E[\beta^{T^*}]\) Divide both sides by \(a\), we have: \(1 \geq E[\beta^{T^*}]\) Since \(0 < \beta < 1\), it must be the case that \(a = E\left[\beta^{T^*} M_{T^*}\right]\). Finally, since \(T^*\) is the first time when the random variable \(Y_n\) reaches or exceeds \(a\), it follows that \(T^*\) maximizes \(E\left[\beta^T M_T\right]\) over all Markov times \(T\).