Prob 11 Page 363 from “Intro to Probability Book”

Given that Xn = Yn+1 - Yn

And that \(\mu\) = 0, \(\sigma\) = 1/2 and Y1 = 100

In 365 days, Yn+1={100,Y2,Y3,….,Y365} and Xn+1 - Xn is (Yn+2-Yn+1) - (Yn+1 - Yn) … => Yn+2 - Yn ….

Therefore, Sum of 365 days of price differentials:

S365=(Y2−100)+(Y3−Y2)+(Y4−Y3)+….+(Y365−Y364)

which results to:

S365 = Y365 - 100, because all the in between terms get cancelled out except the last and the first values

Over 365 days:

Sum of 365 (\(\mu\)) : \(\mu\) + \(\mu\) + \(\mu\) +….= 0 + 0 + 0 + 0 + …..= 0

Sum of 365 Variance: \(\sigma^{2}\) + \(\sigma^{2}\) + \(\sigma^{2}\) + ……= 1/4 + 1/4 + 1/4 + …..= 365*(1/4) = 91.25

Std Deviation: \(\sigma\) = 9.55

standardizing via z score:

z = (point estimate - mu)/Std Dev.

#Part (a)

P(|Y365 - 100| >= 100)

P(|Y365|) >= (0-0)/9.55 = 0, with a normal distribution of $\mu$=0, => P(|Y365|) = 0.5, see R code below

# (a) >= 100

1- pnorm(0)
## [1] 0.5
#Part (b)
P(|Y365 - 100| >= 110)

P(|Y365| >= (110 - 100)/9.55 = 1.046

Z = 1.046 => 1- 0.852 = 0.1477 or 14.8%, see R code below

# (b) >= 110
1- pnorm(1.046)
## [1] 0.1477805
#Part (c)
P(|Y365 - 100| >= 120)

P(|Y365| >= (120 - 110)/9.55 = 2.0942

Z = 2.0942 => 1- 0.9816 = 0.0181 or 1.812%, see R code below

# (c) >= 120
1- pnorm(2.0942 )
## [1] 0.01812108
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