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A model for the movement of a stock supposes that if the present price of the stock is s, then after one time period it will be either (1.012)s with probability 0.52, or (0.99)s with probability 0.48. Assuming that successive movements are independent, approximate the probability that the stock s price will be up at least 30% after the next 1000 time periods.

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  • $\begingroup$ Welcome to MSE! Do you have any thoughts on the problem or tried any approaches and are confused somewhere? Regards $\endgroup$
    – Amzoti
    Apr 9, 2013 at 0:11
  • $\begingroup$ thank you for your ınterest.I have no any ıdea about questıon just I need answer $\endgroup$
    – user71653
    Apr 9, 2013 at 0:15

2 Answers 2

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The following is an analysis that requires logarithms. We will use base $e$ ("natural") logarithms. But if you are more comfortable with base $10$, with minor adjustments it can be done that way. Let $X_i$ be the fractional price "increase" from period $i$ to period $i+1$. Let $Y_i=\ln(X_i)$.

Note that with probability $0.52$, we have $Y_i=\ln(1.012)\approx 0.119286$, and with probability $0.48$, we have $Y_i=\ln(0.99)\approx -0.0100503$.

We assume, as directed, that the $X_i$, and hence the $Y_i$, are independent, despite the fact that this is quite implausible.

Calculate the mean $\mu$ and the variance $\sigma^2$ of the $Y_i$. This is a routine calculation.

Let $W$ be the sum $Y_1+Y_2+\cdots+Y_{1000}$. The $W$ is the sum of $1000$ independent identically distributed random variables. The cumulative distribution function of $W$ is well-approximated by the cumulative distribution function of a normal mean $1000\mu$, variance $1000\mu^2$.

We are interested in the probability that the stock price climbs by a factor of at least $30\%$, so by a factor of at least $1.3$.

The required probability is $1$ minus the probability that the price climbs by a factor of less than $1.3$. We concentrate on finding that.

The stock climbs by less than $30\%$ precisely if $W\lt \ln(1.3)\approx 0.2623643$.

So we want to find the probability that a certain normal with known mean and known variance is $\lt 0.2623643$. That's a standard calculation.

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Hint: the final price is determined by the total number of up and down movements, not by their order. How many ups do you need to be up $30\%$ overall? You have a binomial distribution (approximately). What is the mean value of the number of up movements? What is the variance?

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  • $\begingroup$ Actually I have no ıdea about questıon just I have HW that I could not solve :( $\endgroup$
    – user71653
    Apr 9, 2013 at 0:01
  • $\begingroup$ @user71653: so what is the final price if there are 500 up and 500 down movements? 400 and 600? Can you do that? Then how many ups to you expect out of 1000? $\endgroup$ Apr 9, 2013 at 0:04
  • $\begingroup$ nope still do not understand what you mean. no idea about distributions $\endgroup$
    – user71653
    Apr 9, 2013 at 0:07

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