# Markov process

#### Poirot

##### Banned
It is widely believed that the daily change in currency exchange rates is a random variable with mean 0 and variance v​
That is, if Yn represents the exchange rate on the nth day, Yn = Yn1 + Xn, n = 1, 2, . . . where X1,X2, . . . are independent and identically
distributed normal random variables with mean 0 and variance v. Suppose that today’s exchange rate is 1.55 and v = 0.0025, then what is the probability that the exchange rate will be below 1.4 in 9 days and in 25 days?

Am I meant to setup a diffferential equation, I'm not sure?

#### CaptainBlack

##### Well-known member
It is widely believed that the daily change in currency exchange rates is a random variable with mean 0 and variance v​
That is, if Yn represents the exchange rate on the nth day, Yn = Yn1 + Xn, n = 1, 2, . . . where X1,X2, . . . are independent and identically
distributed normal random variables with mean 0 and variance v. Suppose that today’s exchange rate is 1.55 and v= 0.0025, then what is the probability that the exchange rate will be below 1.4 in 9 days and in 25 days?

Am I meant to setup a diffferential equation, I'm not sure?
The change over $$n$$ days is the sum of $$n$$ iid RV's and so the mean change is $$n$$ times the mean change, in this case $$0$$ and the variance of the change is $$n$$ times the single day variance, so in this case is $$n.v$$

CB

#### Poirot

##### Banned
Thanks CaptainBlack, but I am looking for a probability. If we let Z denote the change in exchange rate over n days, so Z-normal(0,nv), do I compute probability by normal table?

#### CaptainBlack

##### Well-known member
Thanks CaptainBlack, but I am looking for a probability. If we let Z denote the change in exchange rate over n days, so Z-normal(0,nv), do I compute probability by normal table?
The distribution is N(0,nv) and you use probability tables for the computations.

CB