SERVING THE QUANTITATIVE FINANCE COMMUNITY

Alan
Topic Author
Posts: 9915
Joined: December 19th, 2001, 4:01 am
Location: California
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### Catch my error

This puzzle is based upon the setup in this thread. Suppose you repeatedly reinvest all your wealth in a bet where the simple return each time is

$R^+ = 0.675$ with probability 1/2
$R^- = -0.45$ with probability 1/2

In other words, if you start period n with wealth $W_n$, then $W_{n+1} = W_n (1 + R)$, where the R's are independent draws each time from the above distribution.

I wanted to calculate the exact expected utility of a risk-neutral investor whose expected utility over N trials is $\bar{U}_N = E_0 \left(\frac{W_N}{W_0}\right)$.

Here was my thinking.

First, $\bar{U}_N = E_0[e^{X_N}]$, where $X_N = \sum_{n=1}^{N} x_n$ and $x_n = \log(1+R_n)$.
Again, the $R_n$ are i.i.d draws from the simple distribution above.

By the Central Limit Theorem, since the $x_n$ are i.i.d, we expect $X_N$ to tend to a normal distribution with mean $N \mu$ and variance $N \sigma^2$, with the approximation becoming better and better as $N \rightarrow \infty$.

Next, $\mu = E[x_n] = E[\log(1 + R)] \approx -0.041012$, and
$\sigma^2 = Var[x_n] \approx 0.3100541$

Since $X_N$ is normally distributed for large enough N, we can use the EXACT normal relation (or wikipedia log-normal mean, if you like):

(*) $\bar{U}_N = E_0[e^{X_N}] = \exp \{(\mu + 0.5 \, \sigma^2)N \} \approx (1.120769)^N$, using the values just given for $(\mu,\sigma^2)$.

But (*) can't be correct, since the returns are independent each time, and so we must have

(**) $\bar{U}_N = \left( E[(1 + R)] \right)^N = (1.1125)^N$, which contradicts (*).

The brainteaser: where was my mistake?

p.s. Just to be clear, everything I said in that linked thread was correct. The mistake you want to find is in the calculation here. And, yes, I know where it is -- pretty sure

Mars
Posts: 103
Joined: November 13th, 2002, 5:10 pm

### Re: Catch my error

I will try: with CLT you can say that $\frac{ X_N - N \mu }{ \sigma \sqrt{N}}$ tend in law to a normal distribution $\cal{N} (0, 1)$ not that $X_N$ tend in law to  $\cal{N}$ $(N \mu, \sigma \sqrt{N})$.
.

Alan
Topic Author
Posts: 9915
Joined: December 19th, 2001, 4:01 am
Location: California
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### Re: Catch my error

Exactly! Thanks for playing -- I thought it was a lost cause.

katastrofa
Posts: 8625
Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

### Re: Catch my error

I was also wondering what you meant by $\infty \cdot \mu$, but was afraid to ask