Solve X-(S+P-C)E^RT+D*E^RT=0 and Isolate C

  • Thread starter qtiphife
  • Start date
In summary, the equation X-(S+P-C)E^RT+D*E^RT=0 represents a stock's value with dividend due to put-call parity. To isolate C, the equation can be transformed to C = x e^{-Rt} + D e^{R(T-t)} + S - P, with the negative sign corrected and the values for each variable given.
  • #1
qtiphife
5
0
X-(S+P-C)E^RT+D*E^RT=0

How can I "isolate" C?

Thanks
 
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  • #2
Assuming E^RT means E^(RT)...

X - (S + P - C)E^(RT) + D*E^(RT) = 0
<=>
X + D*E^(RT) = (S + P - C)E^(RT)
<=>
X/E^(RT) + D = (S + P - C).

Etc.
 
  • #3
qtiphife,

Please post your questions only once. Triple posting is considered spam.

Thank you,
 
  • #4
So would this be X/E(RT)+D-S+P= C

Man I am terrible with math.
 
  • #5
qtiphife said:
So would this be X/E(RT)+D-S+P= C

Close. It should read:

X/E(RT) + D - S + P = -C

You forgot the negative sign.

Divide both sides by -1, and you're done.
 
  • #6
Oh yeah, and (rt) have different "T's". Thanks.
 
  • #7
TIME OUT!

are you saying that in you original realationship that the T is not the same in the 2 expressions where it is used?

X-(S+P-C)E^RT+D*E^RT=0

Do you actually mean

[tex] x - (s+ p - C) e^{Rt} + D e^{RT} = 0 [/tex]

If so then

[tex] C = x e^{-Rt} + D e^{R(T-t)} - S +P [/tex]
 
  • #8
Guys, thanks a ton for your help, however I am still not getting the correct answer. Here are what the letters stand for:

X (Strike Price)= 50
S (Price of underlying) = 51
P (price of put)= 2.125
e (euler's number) = 2.781
R (rate of interest)= .08
t (time until options expire)= 31/365
T (time until dividend paid)= 25/365
D (dividend)=.46
C (price of call) = ?
I keep on getting a negative number. Any help would be much appreciated. Thanks.
 
Last edited:
  • #9
It would be more helpful if you were to give the physical signifcance of the equation and parameters. Where did you get it? What is it for? Why can't C be negative?
 
  • #10
Sure,

This formula represents a stock's value with dividend due to the put call parity (kind of a pain to explain), but it has to do with buying or selling a stock (underlying) and buying (selling) and selling (buying) options in order to hedge your position in order to create "edge". X represents the strike price the options are bought, and P is the value of the put and C is the value of the call. I need to know the value of the call.

Please look at my previous post for all of the information.

Thanks!
 
Last edited:
  • #11
In your previous posts, you only talk of one T. Now there's two : t and T. So, which is which ?

One more thing. Everyone seems to have copied and repeated a line you wrote, with an uncorrected error :

Post #4 : "So would this be X/E(RT)+D-S+P= C"

Besides, the negative C correction, note that the LHS should really be X/E(RT)+D-S-P. Minus P, not plus.
 
  • #12
Opps! :redface:

You would and should come up with a negitive number, since I lost a negative sign! Try this:

[tex] C =S+P- x e^{-Rt} - D e^{R(T-t)} [/tex]


My apologies for the sloppy algebra!
 

1. How do I solve the equation X-(S+P-C)E^RT+D*E^RT=0 and isolate C?

To solve this equation and isolate C, you will need to use the rules of algebra. First, simplify the equation by combining like terms. Then, isolate C by moving all other terms to the other side of the equation. Finally, solve for C by dividing both sides by the coefficient of C.

2. What is the purpose of isolating C in this equation?

The purpose of isolating C in this equation is to find the specific value of C that makes the equation true. This allows us to solve for the unknown variable and better understand the relationship between the other variables in the equation.

3. Can this equation be solved without isolating C?

Yes, it is possible to solve this equation without isolating C. However, isolating C will provide a more precise and accurate solution.

4. Are there any special techniques or strategies for isolating C in this equation?

There are no special techniques or strategies for isolating C in this equation. It is important to follow the rules of algebra and carefully rearrange the terms in the equation to isolate the variable.

5. How can I check if my solution for C is correct?

To check if your solution for C is correct, you can substitute the value of C back into the original equation and see if it satisfies the equation. You can also use a calculator to evaluate both sides of the equation and see if they are equal.

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