MHB Calculating Probabilities of Exchange Rate Fluctuations with Markov Processes

AI Thread Summary
The discussion focuses on calculating the probabilities of exchange rate fluctuations using Markov processes, specifically under the assumption that daily changes in exchange rates are normally distributed with a mean of 0 and a variance of v. Given an initial exchange rate of 1.55 and a variance of 0.0025, participants are interested in determining the probability of the rate falling below 1.4 in 9 and 25 days. The change in exchange rate over n days is modeled as a normal distribution, Z, with parameters N(0, nv), where nv represents the cumulative variance over those days. To find the probabilities, the normal distribution tables are recommended for calculations. This approach effectively utilizes statistical methods to assess currency exchange rate risks.
Poirot1
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It is widely believed that the daily change in currency exchange rates is a random variable with mean 0 and variance vThat 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?
 
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Poirot said:
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
 
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?
 
Poirot said:
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
 
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