Stock pricing using Binomial model

A stock is prices at $$\100$$ and follows a one-period binomial process with an up move that equals 1.05 and a down move that equals 0.97. If one million Bernoulli trials are performed and the average terminal stock price is $$\102$$, the probability of an up move is closest to ____?

My book gives the solution as follows: $$p \cdot 105 + (1-p) \cdot 97 = 102$$and thus computes the value of $$\text p$$. However. I can't understand how a first step average of the process equals to the average terminal stock price that comes after one million Bernoulli trials. Can someone help me with this question?

• $102=E[S_T]=pS_u+(1-p)S_d$. Then you just plug in the numbers. The “one million” part just means that you have a reasonable estimate of the expectation of the future stock price (after one period). Commented Mar 3, 2020 at 17:45
• Can you please help me understand what does the author mean by average terminal stock price? Is it the stock price after running a million trials pr something else? Commented Mar 3, 2020 at 18:11
• Yes, it is the sample average of 1,000,000 trials. So, it is an estimate for the expected value of the terminal stock price (= stock price in one period). Commented Mar 3, 2020 at 18:19
• Thanks a lot sir. Commented Mar 3, 2020 at 22:58

Assume that you did $$N = 10^6$$ Bernoulli trials. These trials can end up in one of two states up and down.

• In the "up" case the stock is worth \$105. Assume that we have a total of $$U$$ up cases. • In the "down" case the stock is worth \$97. Assume that we have a total of $$D$$ down cases. Clearly $$U + D = N$$.

This means that a proportion of $$\frac{U}{N}$$ of your trials results in an upstate. Relative frequencies are a good approximation for probabilities so you could also say the probability that the trial will result in the upstate is given by $$\frac{U}{N}$$. Define this quantity as $$p$$, i.e. $$p = \frac{U}{N}.$$

Similarly, you can define the probability that your trial will result in the down state as $$\frac D N = \frac {N - U}N = 1-p.$$

Given these probabilities you can compute the expected value: $$e = p \cdot 105 + (1-p) \cdot 97.$$

Now if you know that $$e = 102$$ you can easily solve for $$p$$: $$p = \frac{e - 97}{105 - 97} = \frac 58.$$

The question mentions that 1 million Bernoulli trials are conducted. Suppose the result of $$i^{th}$$ trial is $$X_i$$, then the average of all $$X_i$$s is $$\\\102$$. Thus, we use the weighted average formula.

If the question stated that the average was $$\\\102$$ AFTER 1 million trials, then the following approach needs to be adopted

Probability of $$n$$ successes in $$N$$ trials is $${N\choose n}p^n(1-p)^{N-n}$$. Thus, the final value of the stock price can be $$100*{N\choose n}1.05^n*p^n*0.97^{N-n}(1-p)^{N-n}$$ assuming $$n$$ up moves and (thus) $$N-n$$ down moves.

Therefore, the average stock price should be $$\sum_{n=0}^N100*{N\choose n}1.05^n*p^n*0.97^{N-n}(1-p)^{N-n}=102$$

In other words: $$\sum_{n=0}^N{N\choose n}1.05^n*p^n*0.97^{N-n}(1-p)^{N-n}=1.02$$ where $$N=1,000,000$$

This simplifies to: $$\sum_{n=0}^N{N\choose n}(1.05p)^n*(0.97-0.97p)^{N-n}=1.02$$

Comparing with the formula of binomial expansion

$$(1.05p+0.97-0.97p)^N=1.02$$

Since $$N$$ is large, we get: $$(0.97+0.08p)\approx1$$, which gives $$p=0.375$$