# Binomail Trees for Options - Schweser Practice Exam 1 Morning Session Question 37

I cam across this questions and I don’t know if it’s a problem in the Schweser notes or if it’s a problem with the question.

This is for the morning session of Exam 1 for the Schweser practice tests (2013), question number 37.

“The value of a two period 40 call on Arbor Industries stock is closest to:”

a. 6.65

b. 8.86

c. 9.21

Stock is currently trading at \$45, up move factor is 1.15, risk free rate is 4%. They say the answer is B because of this explanation:

Two upmoves = 45*(1.15^2) = 59.51

Two down moves = 45*(1/1.15)^2 = 34.03

Intrinsic call value worth \$19.51 with two up moves, 0 with two down moves.

Up and a down move the stock price will be at 45. Intrinsic call value is \$5.

Risk neutral probabilities are .607 for up and .393 for down.

I agree with everything up until here.

They continue:

Probabilityy weighted present value of the option pay off for two up moves is \$6.65 because .607^2(19.51)/(1.04^2) gives you \$6.65. I agree with this.

Then they go on to say that "for up-down and down-up (which are equal probabilities, the probability weighted present value of the payoff is (2)(.607)(.393)(\$5)/(1.04^2) = \$2.21 (notice how they DOUBLE the probability of reaching the middle outcome two periods out).

Sum these two answers to get \$8.86

I agree with the logic they are using but if you go back to the schweser notes (LOS 50.b) they do NOT double count the middle result in the binomial tree two periods out. (Example on page 65 in study session 17). By doing the above question how it’s show in the Schweser notes you get C instead of B.

Which way is correct?

I don’t know where you found \ *not double counting the “middle” section but it is never the case. The probability of any end node is always a joint probability, so you always multiply the probabilities to get there. The middle node is times two because there are two paths that lead there.

If you can’t really comprehend any of that stuff a neat trick for the final test :

When you SUM the final probabilities you should be getting one. If you get less or more you are doing something wrong.

Case in point : UP : 60% , DOWN : 40%

Possibilities:

UP UP = 0.6 * 0.6 = .36

UP DOWN = 0.6 * 0.4 = .24

OR 2 *( 0.6*0.4) = .48

DOWN UP = 0.4 *0 .6 = .24

DOWN DOWN = 0.4*.04 =.16

Are they adding up to one? : .36 + .24 + .24 + .16 = 1 Good to go!

I agree with that logic but why aren’t they doing that in the CFAI text or in the Schweser notes? For example in Volume 6 of the CFAI text page 176 Example 5. If you use their method in the question I posted you get C instead of B. They are simply discounting the values to the prior node.

Sorry if you do it the way they have it in the CFAI text you get 6.1341; this is exactly how the CFAI calculated Example 5 on page 176 of Volume 6 - I’m just substituting the numbers with the Schweser mock exam #1 Morning session #37 question:

So U=1.15; (given)

D= .8696; (calculated by 1/U)

ProbU = .6077; (1+Rf-D)/U-D ; Rf = .04

ProbD = .3923; (1-ProbU)

W/ Stock price starting at 45 you get in each period:

S+ = 51.75

S- = 39.132

S++ = 51.5125

S± = 45

S- - = 34.03

Then find the option prices at expiration:

C++ = Max(0,51.5125-40) = 11.5125

C± = Max (0,45-40) = 5

C-- = Max (0, 34.03-40) = 0

Taking the value of the ProbU and ProbD:

Step back and find the option prices at time 1:

C+= (11.51)(.6077)+(5)(.3929)/1.04 = 8.6117

C- = (5)(.6077) + (0)(.3929)/1.04 = 2.9216

The price today is:

C= 8.6117(.6077)+2.9216(.3929)/1.04 = 6.1341

This is how both Schweser Notes and CFAI solves the problem. Why is Schweser Mock solving it as per my first post.

Ok, having seen the example you mentioned :

CFAi does not use joint probabilities but instead works out the cost at each step. If you decompose the answer you should notice that the value (.44) is used twice because we can arrive there with two ways. Either UP DOWN, or DOWN UP.

I personally feel that doing the joint probabilities is much faster and easier. To solve the problem again go like this :

for the UP UP = (.56^2) * \$8.99 = \$2.81926

for the DOWN UP and UP DOWN = (2*.56*.44) * .44 = .21683

for the DOWN DOWN = (.44^2)*\$0 = 0

First notice that the probabilities (bold) add up to one = .3136 + .4928 + .1936 = 1

Second, bear in mind that the values are two periods in the future so we will need to discount accordingly:

(2.81926 + .21683)/(1.03^2) ~= \$2.86

Both methods double count the middle node. Use whichever method seems easier to you!

Verse I did this exercise (I did the mock yesterday). I can guarantee than if there are two periods, the middle node will be accounted for twice. Rework the example slowly and carefully and you will see it yourself. If you are still struggling I’d be glad to work it step by step with you either method Ok - I see what you’re saying - essentially both ways are the same thing.

But if this is the case - why am I not getting the same answer (B) that Schweser is providing in my original post?

What am I doing wrong in my post showing all of the calculations?

I am in the middle of a phonecall but I will look at it soon!

Great thanks man

NM - i’m sorry for all the trouble. I see that i’m an idiot and miscalculated the C++

Totally works now. Thanks again!

Not a problem man! Good luck with the rest.