Stock pricing using Binomial model

Question

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?

Answer 1

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 $p$: $$ p = \frac{e - 97}{105 - 97} = \frac 58. $$