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Thread: maximize a sequence of functions

  1. #1
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    maximize a sequence of functions

    Hallo,

    Someone already helped me to show that the following convergence is uniformly for all $\displaystyle a \in [0,1]$:
    $\displaystyle \lim\limits_{h \to 0}\left(1+a(\mu-r)h+\frac{1}{2}a^2 \gamma(1-\gamma)\sigma^2h+O(h^\frac{3}{2})\right)^\frac{1}{ h}=e^{a(\mu-r)+ \frac{1}{2}a^2 \gamma(1-\gamma)\sigma^2}.$

    whereas
    $\displaystyle O(h^\frac{3}{2})$ is the remainder term of a taylor expansion where all terms with factor $\displaystyle h^\frac{3}{2}$ and higher exponents are collected. This term can be bounded by a constant that can be chosen independent of $\displaystyle a$
    $\displaystyle \mu \in \mathbb{R}$ is a constants
    $\displaystyle \sigma,r >0 $ are constants
    $\displaystyle 0<\gamma <1 $ is a constant
    $\displaystyle f_h(a):=\left(1+a(\mu-r)h+\frac{1}{2}a^2 \gamma(1-\gamma)\sigma^2h+Ch^\frac{3}{2})\right)^\frac{1}{h }$
    $\displaystyle f(a):=e^{a(\mu-r)+ \frac{1}{2}a^2 \gamma(1-\gamma)\sigma^2}$


    does the maximum point of the sequence of functions $\displaystyle (f_h(a))_{h \in (0,1)}$ (I call it $\displaystyle x_h$) with $\displaystyle a \in [0,1]$ converge to the maximum point of $\displaystyle f$ for (I call it $\displaystyle x$) $\displaystyle h\to 0$?


    Does anyone know if the following holds:
    $\displaystyle \lim\limits_{h \to 0}x_h = x$?



    I only know a theorem which says that this is true for a sequence of concave functions. But my functions are neither concave nor convex for all $\displaystyle a \in [0,1]$.

    Can anybody help me?
    Thanks!
    Last edited by Juju; Mar 18th 2011 at 01:29 PM.
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  2. #2
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    I know that $\displaystyle (f_h(a))^h$ is concave in $\displaystyle a$. So there's a unique maximum point of $\displaystyle (f_h(a))^h$ on $\displaystyle [0,1]$.
    Since the function $\displaystyle x^\frac{1}{h}, x\geq0$ is strictly increasing, $\displaystyle (f_h(a))^h$ and $\displaystyle f_h(a)$ have the same maximum point.

    Can I now conclude that $\displaystyle f_h(a)$ has also a unique maximum point?

    Also $\displaystyle \log_e(f(a))$ is concave in $\displaystyle a$ and $\displaystyle e^x$ is strictly increasing.
    Then the maximum point of $\displaystyle f(a)$ on $\displaystyle [0,1]$ is unique. Am I right?

    If the maximum points of $\displaystyle f_h$ and $\displaystyle f$ are unique I think it follows from the uniform convergence that it holds
    $\displaystyle x_h \to x$.
    Am I right?

    Sorry, I made some mistakes in the definitions. There should be a $\displaystyle (\gamma-1)$ instead of the $\displaystyle (1-\gamma)$. So, the corrected version:
    $\displaystyle f_h(a):=\left(1+a(\mu-r)h+\frac{1}{2}a^2 \gamma(\gamma-1)\sigma^2h+O(h^\frac{3}{2})\right)^\frac{1}{h}$
    $\displaystyle f(a):=e^{a(\mu-r)+ \frac{1}{2}a^2 \gamma(\gamma-1)\sigma^2}$
    Last edited by Juju; Mar 19th 2011 at 12:38 PM.
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