Many people believe that the daily change of price of a company’s stock on the stock market is a random variable with mean 0 and variance 2. That is, if Y represents the price of the stock on the nth day, then Y n = Y n − 1 + X n n ≥ 1 where X 1 , X 2 , ... are independent and identically distributed random variables with mean 0 and variance σ 2 . Suppose that the stock’s price today is 100. If σ 2 = 1 , what can you say about the probability that the stock’s price will exceed 105 after 10 days?
Many people believe that the daily change of price of a company’s stock on the stock market is a random variable with mean 0 and variance 2. That is, if Y represents the price of the stock on the nth day, then Y n = Y n − 1 + X n n ≥ 1 where X 1 , X 2 , ... are independent and identically distributed random variables with mean 0 and variance σ 2 . Suppose that the stock’s price today is 100. If σ 2 = 1 , what can you say about the probability that the stock’s price will exceed 105 after 10 days?
Solution Summary: The author explains the value of the probability that the stock price exceeds 105 after 10 days.
Many people believe that the daily change of price of a company’s stock on the stock market is a random variable with mean 0 and variance 2. That is, if Y represents the price of the stock on the nth day, then
Y
n
=
Y
n
−
1
+
X
n
n
≥
1
where
X
1
,
X
2
,
...
are independent and identically distributed random variables with mean 0 and variance
σ
2
. Suppose that the stock’s price today is 100. If
σ
2
=
1
, what can you say about the probability that the stock’s price will exceed 105 after 10 days?
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