probability of an up move in the stock

A stock is priced at 40 and the periodic risk-free rate of interest is 8%. The value of a two-period European call option with a strike price of 37 on a share of stock using a binomial model with an up factor of 1.20 and down factor of 0.833 is closest to:

Answer:

First, calculate the probability of an up move or a down move:

Pu = (1 + 0.08 − 0.833) / (1.20 − 0.833) = 0.673 Pd = 1 − 0.673 = 0.327

Two up moves produce a stock price of 40 × 1.44 = 57.60 and a call value at the end of two periods of 20.60. An up and a down move leave the stock price unchanged at 40 and produce a call value of 3. Two down moves result in the option being out of the money. The value of the call option is discounted back one year and then discounted back again to today. The calculations are as follows:

C+ = [20.6(0.673) + 3(0.327)] / 1.08 = 13.745

C- = [3(0.673) + 0 (0.327)] / 1.08 = 1.869

Call value today = [13.745(0.673) + 1.869(0.327)] / 1.08 = 9.13

…I completely get the valuation part of this but I can’t seem to conceptually understand why we can calculate the probability of an up move as shown above??

It’s _ not _ a probability.

They call it a probability, but it isn’t. It has nothing to do with how likely it is that the price of the stock will go up.

It’s a weight. Nothing more, nothing less.

It’s the weight that results in the weighted average future price being today’s price increased at the risk-free rate. You can set up the formula for the weighted average future price, then set it equal to today’s price times (1 + rf), then solve for the weight on the up price.

Technically, the weights (for an up move and a down move) are called the risk-neutral probabilities , but they’re really nothing more nor less than weights.

still having trouble making ANY sense of why its used then…so if the down factor was .5 instead of .83, that means the weight asgined to an up move is greater? any logic here?

sorry- def know i’m missing something important here…

If you wan to think of the weights as probabilities, that’s OK. Ask yourself which is more likely: a price drop of 17% or a price drop of 50%.

Therefore, the weight on a 50% drop would be less than the weight on a 17% drop, making the weight on an increase greater when the down factor is 0.50 than when it’s 0.83.

Actually risk-neutral probabilities are real probabilities.

Let call p the probability of an up move and 1-p the probability of a down move.

p >= 0 and p + (1-p) = 1 so we have a proper discrete probability distribution over the universe Up and Down.

The idea behind risk neutral probability is that we use the risk-free rate to discount the value of an asset and not a discount rate that reflects normal risk aversion.

Let S0 be the initial stock price, S*U the up state value and S* D the down value.

S0 = (SUp + (1-p)SD)/(1 + RF) .

SUp + SD -pSD = S(1+RF)

We divide by S both sides

Up + D -pD = 1 + RF

p(U-D) = 1 + RF - D

p = (1 + RF - D)/(U-D)

They’re not probabilities in the sense that they represent the expectation or likelihood that something will happen.

They’re simply weights that are nonnegative and sum to one. Those are the only characteristics that they have in common with probabilities.

Let’s use an example. So = 30, U = 1.2 and D = 0.8 and RF = 0.03

P = (1.03 - 0.8)/(1.2-0.8) = 0.575 and SU = 36 and SD = 24

I agree with you that the probability that in one period the value of S will be 36 is probably not 0.575 in real life.

But, if we agree with the assumptions of the binomial model where the stock can only be 36 or 24 than yes, the probability of the stock being 36 is 0.575 and the probability of the stock being 24 is 0.425

No . . . if we agree with the assumptions of the binomial model where the stock (price) can only be 36 or 24, then the weight we have to assign to 36 has to be 0.575 and the weight we have to assign to 24 has to be 0.425 so that the tree is arbitrage-free.

Seriously: they’re not probabilities; they’re just weights. There’s nothing wrong with them being just weights, but that’s all they are.

OK. What is the rationale for the weights? I we consider we have probabilities than everything makes sense: discounted expected value.

The rationale is that these are the weights needed to have an arbitrage-free tree.

Discounted weighted-average value; what’s wrong with that?

(Note: I, personally, don’t have any problem with calling them probabilities. But many of the candidates whom I have taught and tutored have had difficulties with that view (“Why is there a 57.5% chance that the price will go up to 36? How do we know that?”), but no difficulty with the view that they’re merely weights assigned to achieve a specific goal (no arbitrage). I tend to go with what works for the majority of candidates.)

by the way thanks a lot I think I have a much better understanding now…

this comment is very helpful I think: ‘if we agree with the assumptions of the binomial model where the stock (price) can only be 36 or 24, then the weight we have to assign to 36 has to be 0.575 and the weight we have to assign to 24 has to be 0.425 so that the tree is arbitrage-free’

Quite fine. Thank you for taking the time for this discussion.

My pleasure.

By the way, as a subtle point (in real life, not on the CFA exam), we would have potentially different risk-free rates for each maturity, so we would have different up-move and down-move weights for each level of the tree.

On the exam, of course, we’ll have the same risk-free rate for every maturity, so you calculate the weights once for all.

Interesting point. Thank you