### Catch my error

Posted:

**December 14th, 2018, 10:10 pm**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

[$]R^+ = 0.675[$] 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