Expected value and variance question on asset prices, 95% confidence interval.

Question

You have $100 dollars. For 100 days, it either goes up or down by 10% randomly each day. What is the 95% confidence interval in which this value will lie in 100 days?

My idea:

I tried to take advantage of variances and standard deviations and thought that the daily variance would be .01 and I would multiply this value by 200 to get the 95% confidence interval and then add and subtract this value to 100. However, I think this is very wrong and was hoping someone could give me some insight on how to best go about solving this problem.

Additionally, I tried to branch out the structure for the first two days to see if that revealed anything and found that we ended up at 121 a fourth of the time, 99 half of the time, and 81 a fourth of the time and then tried to find the variance which came out to be the sqrt(201) but I did not know how to relate this to 100 days.

Answer 1

The problem is that the variance of the asset price varies day by day depending on its level, which can not be scaled up directly. Introduced below is a transformed variable whose variance can be proportionally scaled up.

Let $X$ be a symmetric (or normal) random variable starting at 0 and with daily change of either up amount $a$ or down amount $a$. And, let the dollar price $Y$ related to $X$ via

$$Y_n=100\exp\left( X_n - \frac 12 a^2t_n \right)$$

where $n$ denotes number of days and $t_n$ is number of days. The above relationship ensures that the expected price of $Y$ is 100.

In order to have $Y$ going up or down 10% each day, the daily change $a$ for $X$ has to be

$$a= \ln\frac{1.1}{0.9}\approx 0.1003353 $$

Note that $a^2$ is the daily variance of $X$ and its variance in 100 days is $100a^2$. The two standard deviations, or of the 95% confidence, is $2\sqrt{100a^2}=20a$.

Therefore, the price in 100 days lies within

$$ 100\exp\left( -20a - \frac 12 a^2 100 \right) < Y_{100} < 100\exp\left( 20a - \frac 12 a^2 100 \right) $$

which is,

$$ 8.13< Y_{100} < 449.7$$

Answer 2

You are correct to try to use the variance. Here is how you can proceed:-

Let $X_n$ be the amount after $n$ days with $X_0=100.$

Then $X_{n+1}$ is $1.1X_n$ with probability $0.5$ and $0.9X_n$ with probability $0.5$.

Therefore $E(X_{n+1})=0.5E(1.1X_n)+0.5E(0.9X_n)=E(X_n)$, as would be expected. Therefore $E(X_{100})=100$.

More interestingly, $E(X_{n+1}^2)=0.5E(1.21X_n^2)+0.5E(0.81X_n^2)=1.01E(X_n^2)$. Therefore $ E(X_{100}^2)=1.01^{100}E(X_0^2)=1.01^{100}$x$10^4.$

The variance of $X_{100}$ is therefore $(1.01^{100}-1)$x$10^4$.

Note

This method gives the variance after 2 days as $(1.01^2-1)$x$10^4=201$. The s.d. is the square root of this.