0
$\begingroup$

The unit price of a certain commodity evolves randomly from day to day with a general downward drift but with an occasional upward jump when some unforeseen event excites the markets. Long term records suggest that, independently of the past, the daily price increases by a dollar with probability $0.45$, declines by $2$ dollars with probability $0.5$, but jumps up by $10$ dollars with probability $0.05$. Let $C_0$ be the price today and $C_n$ the price $n$ days into the future. How does the probability $P(C_n>C_0)$ behave as $n$ goes to infinity?

My work:

Consider $\lim_{n \to \infty} (C_n>C_0)$ where $Cn=\frac{\Sigma_{i = 1}^{n}X_i}{n}$

where $X_i$ are i.i.d. random variables. Let $\mu$ be the mean of $X_i$ and $\sigma$ be the standard deviation.

$\lim_{n \to \infty} P(C_n>C_0)=1-P(C_n<C_0) <1-P(|C_{n-m}|<C_0) <(1-(1-\frac{sigma^2}{nC_0^2}))=0$

$\endgroup$
  • $\begingroup$ I am not confident with the last part. $\endgroup$ – Michelle Feb 21 at 5:27
  • $\begingroup$ Also, do I compute E(X) and Var(X)? $\endgroup$ – Michelle Feb 21 at 5:31
0
$\begingroup$

Suppose $D_n = C_n - C_0$. Then $P(C_n > C_0) = P(D_n > 0) = P(\frac{D_n}{n} > 0)$, and $D_n = \Sigma_{k = 1}^{n} X_i$, where $X_i$ are i.i.d. random variables, such that $P(X_i = 1) = 0.45$, $P(X_i = -2) = 0.5$ and $P(X_i = 10) = 0.05$. Then by the Law of Large Numbers $\frac{D_n}{n}$ converges in probability to $EX_i = 0.45*1 + 0.5*(-2) + 0.05*10 = -0.05 < 0$. That results in $P(C_n > C_0) = P(\frac{D_n}{n} > 0) = 0$

You can find more about convergence in probability here: https://en.wikipedia.org/wiki/Convergence_of_random_variables#Convergence_in_probability

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.