# How to compute $\lim_{n \to \infty}P(C_n>C_0)$?

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.

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

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

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 $$P(C_n > C_0) = P(\frac{D_n}{n} > 0) = 0$$