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Markov process

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Poirot

Banned
Feb 15, 2012
250
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
Jan 26, 2012
890
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
 
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Poirot

Banned
Feb 15, 2012
250
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
Jan 26, 2012
890
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