Revision as of 01:37, 7 May 2023 by Admin (Created page with "An investor invests 100 dollars in a stock. Each month, the investment has probability 0.5 of increasing by 1.10 dollars and probability 0.5 of decreasing by 0.90 dollars. The...")
May 07'23
Exercise
An investor invests 100 dollars in a stock. Each month, the investment has probability 0.5 of increasing by 1.10 dollars and probability 0.5 of decreasing by 0.90 dollars. The changes in price in different months are mutually independent.
Calculate the probability that the investment has a value greater than 91 dollars at the end of month 100.
- 0.63
- 0.75
- 0.82
- 0.94
- 0.97
May 07'23
Solution: E
Let [math]X_k[/math] be the random change in month [math]k[/math]. Then [math]\operatorname{E}(X_k) = (0.5)(1.1) + 0.5(−0.9) = 0.1[/math] and [math]\operatorname{Var}(X_k) = 0.5(1.1)^2 + 0.5(−0.9)^2 − (0.1)^2 = 1.[/math] Let [math]S = \sum_{k=1}^{100}X_k [/math]. Then, [math]\operatorname{E}(S) = 1000(0.1) = 10[/math] and [math]\operatorname{Var}(S) = 100(1) = 100 [/math]. Finally,
[[math]]
\operatorname{P}(100 + S \gt 91) = \operatorname{P}(S \gt -9) = \operatorname{P}( Z \gt \frac{-9-10}{\sqrt{100}} = -1.9 ) = 0.9713.
[[/math]]