What does this mathematical expression mean?

[larryg003]

The Cox-Rubenstein Model for Calls
is
C = 1/(rr)^n[sum(from k=0, n) of (n!/(k!(n-k)!))(p^k)*(1-p)^(n-k)max(0, u^k*d^(n-k)*U-E]

What does the max(0, u^(k)*(d^(n-k)U-E) mean?

More generally, what does the max() function mean when it follows an infinite series?

Much appreciated,

 

[EdgeHunter]

Quote from larryg003:

The Cox-Rubenstein Model for Calls
is
C = 1/(rr)^n[sum(from k=0, n) of (n!/(k!(n-k)!))(p^k)*(1-p)^(n-k)max(0, u^k*d^(n-k)*U-E]

What does the max(0, u^(k)*(d^(n-k)U-E) mean?

More generally, what does the max() function mean when it follows an infinite series?

Much appreciated,

If no one is succesfully able to answer your question or if its not answered in a timely manner you might want to go...

http://www.wilmott.com/home.cfm

 

[Random.Capital]

Quote from larryg003:

What does the max(0, u^(k)*(d^(n-k)U-E) mean?

It's to ensure the value coming out of max is never negative, ie, can't be less than 0. At expiry, the messiness resolves to (SpotPrice-Strike), so think of the messiness as giving you the value at an arbitrary (binomial tree) node between "now" and "expiry", which also can't be less than 0.

 

[Martinghoul]

Quote from larryg003:
The Cox-Rubenstein Model for Calls
is
C = 1/(rr)^n[sum(from k=0, n) of (n!/(k!(n-k)!))(p^k)*(1-p)^(n-k)max(0, u^k*d^(n-k)*U-E]

What does the max(0, u^(k)*(d^(n-k)U-E) mean?

More generally, what does the max() function mean when it follows an infinite series?

Much appreciated,

I am not familiar with this expression for CRR, but, in general, there's nothing alien about it. The max function is under the Sigma (Sum), so this just means you calculate the max for every value of k between 0 and n.

 

[jimbojim]

Quote from larryg003:

The Cox-Rubenstein Model for Calls
is
C = 1/(rr)^n[sum(from k=0, n) of (n!/(k!(n-k)!))(p^k)*(1-p)^(n-k)max(0, u^k*d^(n-k)*U-E]

What does the max(0, u^(k)*(d^(n-k)U-E) mean?

More generally, what does the max() function mean when it follows an infinite series?

Much appreciated,

It has to do with whether the pricing is for European or American options. The formula is an approximation.

 

[Equalizer]

Quote from larryg003:

The Cox-Rubenstein Model for Calls
is
C = 1/(rr)^n[sum(from k=0, n) of (n!/(k!(n-k)!))(p^k)*(1-p)^(n-k)max(0, u^k*d^(n-k)*U-E]

What does the max(0, u^(k)*(d^(n-k)U-E) mean?

More generally, what does the max() function mean when it follows an infinite series?

Much appreciated,

max(a, b) simply refers to the largest value in the brackets, either a or b, that's all.
You'll notice that some authors use max(a,b) and (a,b)^+ (i.e., the "+" is a superscript) interchangeably.

Setting one of the values (a or b) to zero simply sets the lowest possible value to be non-negative. Note that this refers to intrinsic value only, hence for a simple call it cannot be negative.

 

[Equalizer]

...intrinsic value as in max(0, S - K) that is

 

[larryg003]

Thanks all, especially equalizer.

Now, in your expression you said it could be one or the other?
So it is not a bound like (-3, 5]?

 

[automateddaytra]

Quote from Equalizer:

max(a, b) simply refers to the largest value in the brackets, either a or b, that's all.
You'll notice that some authors use max(a,b) and (a,b)^+ (i.e., the "+" is a superscript) interchangeably.

Setting one of the values (a or b) to zero simply sets the lowest possible value to be non-negative. Note that this refers to intrinsic value only, hence for a simple call it cannot be negative.

You're right, this expression is just to get the max value, because of the complicated expression of the second value, it makes us so incomprehensible

 

[Equalizer]

no it is not a bound (or interval), it simply is the max of the two values. What the zero does is set the lowest value it will return in case S - K (i.e., stock - strike)is negative.

Basically, max(0, S - K) is shorthand for saying that if (for a call) the stock's value is lower than the strike, then the option is worthless (0), otherwise it is S - K. Note max(0, S - K) is intrinsic value only, no time value here.

Think of it as a computer function with two arguments (formal parameters), which returns the larger of the two arguments.