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# Miller - Chapter 4 - Distributions - Study Notes, Pg 75, question #2

#### Dr. Jayanthi Sankaran

##### Well-Known Member
Hi David,

In question (2) referenced as above,

(2) At the start of the year, a stock price is $100. A twelve-step binomial model describes the stock price evolution such that each month the extreme volatility price with either jump from S(t) to S(t)*u with 60.0% probability or down to S(t)*d with 40.0% probability. The up jump (u) = 1.1 and the down jump (d) = 1/1.1; note these (u) and (d) parameters correspond to an annual volatility of about 33% as exp[33%*SQRT(1/12)] ~= 1.10. At the end of the year, which is nearest to the probability that the stock price will be exactly$121?

(a) 0.33%
(b) 3.49%
(c) 12.25%
(d) 22.70%

There are 13 outcomes at the end of the 12-step binomial, with $100 as the outcome that must correspond to six up jumps and six down jumps. Therefore,$121 must be the outcome due to seven up jumps and five down jumps: $100*1.1^7*(1/1.1)^5 =$121
Such that we want the binomial probability given by:
Binomial Prob [X = 7| n =12, p = 60%] = 22.70%

Clarification

Although, intuitively, 13 outcomes at the end of the 12-step binomial seems right (ie expected value of stock price at the end of 12 steps ??), how do you figure out six up jumps and six down jumps from the Binomial model to get \$100 as the outcome? Also how do you get :
Binomial Prob [X = 7| n =12, p = 60%] = 22.70%
I guess I am missing something that I should understand.

Thanks!
Jayanthi

#### ShaktiRathore

##### Well-Known Member
Subscriber
Hi
Consider n up jumps and 12-n down jumps to get 121 in the end
Therefore,100*(1.1)^n*(1/1.1)^12-n=121=>1.1^(2n-12)=1.1^2=>2n-12=2=>2n=14=>n=7 therefore there are 7 up and 12-7=5 down jumps.
Binomial probability is 12C7(.6)^7*(.4)^5=22.69%
Thanks

#### Dr. Jayanthi Sankaran

##### Well-Known Member
Hi Shakti,

Yes - that is indeed great! Thanks a tonne for making it so simple to understand!

Jayanthi

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @Jayanthi Sankaran the source question thread contains a visual that should help clarify, too. Please see https://www.bionicturtle.com/forum/...ributions-i-miller-chapter-4.7025/#post-28837
ie..,

#### Dr. Jayanthi Sankaran

##### Well-Known Member
This is neat, David

Thanks!
Jayanthi

#### Dr. Jayanthi Sankaran

##### Well-Known Member
Hi David,

Does the above visual come with its own spreadsheet?

Thanks
Jayanthi

#### Dr. Jayanthi Sankaran

##### Well-Known Member
Thanks David - will go through it, at leisure

Warm regards
Jayanthi

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @eva maria That wasn't me, as explained i think in the linked discussion, I tend to prefer here a visual/intuitive approach (which is a matter of style). Shakti's solution is mathematically elegant, he didn't strictly type his parens but to be honest I think it helps to go with the flow sometimes and replicate what somebody is trying to show you, when math is involved. What I mean is that, he's quickly shared an elegant solution such that, sometimes the best thing to do is replicate the steps:
• 100*(1.1)^n*(1/1.1)^(12-n) = 121; i.e., the fundamental assumption. Then he divides each side by 100:
• 1*(1.1)^n*(1/1.1)^(12-n) = 1.21; then realize that (1/1.1) = 1.1^(-1) such that we have:
• 1*(1.1)^n*[1.1^(-1)]^(12-n) = 1.21 -->
• (1.1)^n * 1.1^[-1 * (12-n)] = 1.21 --> because (a^x)^y = a^(x*y), such that:
• (1.1)^n * 1.1^(n-12) = 1.21,
• (1.1)^[n+(n-12)] = 1.21, because a^x * a^y = a^(x+y), and
• (1.1)^(2n - 12) = 1.21 = 1.1^2; he's really skilled so he probably did all of that in his head and didn't worry about expressing every little detail. Whereas I am actually not so quick and i need to detail it all out. I hope that explains, thanks!

#### eva maria

##### New Member
Hi @eva maria That wasn't me, as explained i think in the linked discussion, I tend to prefer here a visual/intuitive approach (which is a matter of style). Shakti's solution is mathematically elegant, he didn't strictly type his parens but to be honest I think it helps to go with the flow sometimes and replicate what somebody is trying to show you, when math is involved. What I mean is that, he's quickly shared an elegant solution such that, sometimes the best thing to do is replicate the steps:
• 100*(1.1)^n*(1/1.1)^(12-n) = 121; i.e., the fundamental assumption. Then he divides each side by 100:
• 1*(1.1)^n*(1/1.1)^(12-n) = 1.21; then realize that (1/1.1) = 1.1^(-1) such that we have:
• 1*(1.1)^n*[1.1^(-1)]^(12-n) = 1.21 -->
• (1.1)^n * 1.1^[-1 * (12-n)] = 1.21 --> because (a^x)^y = a^(x*y), such that:
• (1.1)^n * 1.1^(n-12) = 1.21,
• (1.1)^[n+(n-12)] = 1.21, because a^x * a^y = a^(x+y), and
• (1.1)^(2n - 12) = 1.21 = 1.1^2; he's really skilled so he probably did all of that in his head and didn't worry about expressing every little detail. Whereas I am actually not so quick and i need to detail it all out. I hope that explains, thanks!

Many thanks David

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