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?

  • 2
    $\begingroup$ $102=E[S_T]=pS_u+(1-p)S_d$. 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). $\endgroup$
    – Kevin
    Commented Mar 3, 2020 at 17:45
  • $\begingroup$ 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? $\endgroup$ Commented Mar 3, 2020 at 18:11
  • 1
    $\begingroup$ 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). $\endgroup$
    – Kevin
    Commented Mar 3, 2020 at 18:19
  • $\begingroup$ Thanks a lot sir. $\endgroup$ Commented Mar 3, 2020 at 22:58

2 Answers 2


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


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


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