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Thread: Help w/ substituting, rearranging formula: V = S^(1/1) e^(0.4(1-0)) = S e^(2/5)

  1. #1

    Help w/ substituting, rearranging formula: V = S^(1/1) e^(0.4(1-0)) = S e^(2/5)

    Dear all,

    I'm running into the following problem and I would like to have some assistance since I do not understand how to inverse a calculation and substitute it into an equation.

    Assume the following: V = S(1/1)e0.4(1-0) --> e(2/5)S and the the first derivative is e(2/5)

    Now I have to fill in an equation with the first derivative which looks like:
    dv = (e(2/5)*(0.14-0.01)S)

    The 0.14 and 0.01 are given.

    Then the question is to substitute S in formula dv with the inverse of e(2/5)
    1. Can someone explain what the inverse of e(2/5) is? I thought LN(2/5)?
    2. How I substitute the inverse of e(2/5) into the dv equation for S? (I know the S will be replaced with the inverse but the formula should also be simplified or rearranged whatever that is called.

    If someone can explain the above two questions I would be very thankful. Otherwise it will take me days since I do not have a background in even te basics of math

  2. #2
    I think it helps if I clarify the context of the question.
    The question is about valuing a derivative using ito process and ito lemma.

    For instance, alpha (a) is 0.14, delta (s) is 0.01 and sigma (o) is 0.48

    Formula ito lemma: (VS(a-d)S + Vt + 1/2Vss(oS)2 ) Dt + VsoSdz

    Vs = first derivative, Vss is second derivative Vt is first derivative with respect to time.

    The value of the derivative can be described by: V = S(1/1)e0.4*(1-t)
    I think that Vs = e0.4s and the second derivative is 0.

    Then the formula becomes:
    (e0.4S(0.14-0.1)S + Vt(???) + 1/2*)*(0.48S)2 ) Dt + e0.4s*0.48S dz
    (e0.4S(0.14-0.1)S + Vt(???))Dt + e0.4s*0.48S dz

    Questions:
    1. S has to be substituted by the inverse of V = S(1/1)e0.4*(1-t) but I don't know what the inverse is;
    2. I don't know what to do with t or how to find the first derivative with respect to t.
    3. How to substitute the inverse in the equation finally?

  3. #3
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by nasdreas View Post
    If someone can explain the...two questions I would be very thankful. Otherwise it will take me days since I do not have a background in even the basics of math
    Why are you being assigned homework involving algebra and calculus when you haven't yet taken those courses?

    Quote Originally Posted by nasdreas View Post
    Assume the following: V = S(1/1)e0.4(1-0) --> e(2/5)S and the the first derivative is e(2/5)
    So V is a function of S? And the derivative will be V'(S) = dV/dS?

    Quote Originally Posted by nasdreas View Post
    Now I have to fill in an equation with the first derivative which looks like:
    dv = (e(2/5)*(0.14-0.01)S)
    What do you mean by "filling in an equation"? How does "v" relate to "V", if at all? With respect to what variable is the derivative taken (that is, what is underneath the "dv")?

    Quote Originally Posted by nasdreas View Post
    Then the question is to substitute S in formula dv with the inverse of e(2/5)
    Are these maybe the "instructions"...? (If "substitute S in formula dv with the inverse of e(2/5)" indicates "the question", then I'm not understanding the question.)

    Quote Originally Posted by nasdreas View Post
    1. Can someone explain what the inverse of e(2/5) is? I thought LN(2/5)?
    Lacking a context (additive inverse, multiplicative inverse, functional inverse if we're saying "y = e^x; find inverse statement, given x = 2/5", etc), I'm not sure.

    Quote Originally Posted by nasdreas View Post
    2. How I substitute the inverse of e(2/5) into the dv equation for S?
    No idea. See above.

    Quote Originally Posted by nasdreas View Post
    ...the context of the question...is...valuing a derivative using ito process and ito lemma.

    ito lemma: (VS(a-d)S + Vt + 1/2Vss(oS)2 ) Dt + VsoSdz

    Vs = first derivative, Vss is second derivative Vt is first derivative with respect to time.
    What is the "ito process"? Does "Vs" mean "the first derivative of V with respect to s, or with respect to S, or with respect [see below] to [tex]\delta[/tex])? (Similar question for "Vss".)

    Quote Originally Posted by nasdreas View Post
    For instance, alpha (a) is 0.14, delta (s) is 0.01 and sigma (o) is 0.48
    As originally posted, there was no [tex]\alpha,\, \delta,\,[/tex] or [tex]\sigma.[/tex] How do these relate?

    Also, does the above mean that the "ito lemma" is as follows?

    . . . . .[tex]\left(VS\, (\alpha\, -\, d)\, S\, +\, Vt\, +\, \dfrac{1}{2V''\,(\sigma S)^2}\right)\, Dt\, +\, V'\, \sigma Sdz[/tex]

    Quote Originally Posted by nasdreas View Post
    The value of the derivative can be described by: V = S(1/1)e0.4*(1-t)
    Which derivative is this? What do you mean by "the value" of the derivative, when the right-hand side does not actually simplify to a numerical value? What do you mean by "can be described by"?

    Quote Originally Posted by nasdreas View Post
    I think that Vs = e0.4s and the second derivative is 0.
    How did you arrive at this?

    Quote Originally Posted by nasdreas View Post
    Then the formula becomes:
    (e0.4S(0.14-0.1)S + Vt(???) + 1/2*)*(0.48S)2 ) Dt + e0.4s*0.48S dz
    (e0.4S(0.14-0.1)S + Vt(???))Dt + e0.4s*0.48S dz
    Which formula? How?

    Please reply showing all of your work. Thank you!

  4. #4
    Why are you being assigned homework involving algebra and calculus when you haven't yet taken those courses?
    Never had those courses and never had math at this level haha so everything went well until...

    So V is a function of S? And the derivative will be V'(S) = dV/dS?

    The whole idea is as follows. Assuming a stock follows an ito process. Then, a derivative (for instance a call option or put option or any other contract that derives its value from the stock) can be calculated by using the ito lemma.
    Based on that, I think the v is a function of S. The only thing i know is that I have to take the first derivative and second derivative of the formula: V = S(1/1)e0.4(1-t)
    and the first derivative with respect to time. However, T is not given. So im still in doubt whether or not to use a 0 for t or just t in there. But leaving t in these makes is quite hard to find the derivatives.

    What do you mean by "filling in an equation"? How does "v" relate to "V", if at all? With respect to what variable is the derivative taken (that is, what is underneath the "dv")?
    (VS(αd)S+Vt+12V(σS)2)Dt+VσSd
    I have to fill in the above equation. VS is the first derivative of the formula V = S(1/1)e0.4(1-t)
    α = 0.14, d = 0.01, 2V is the second derivative. VT, is the first derivative with respect to time.
    If t = 0 then the first derivative is e0.4s, the second is 0 and the derivative with respect to time is 0 as well. However, I think that T is not zero per definition. The formula just indicates that there is a time component

    So my first questions here are whether or not to take derivative with t being t. And what the derivatives (first, second and time) are? If I know these values, then VS can be replaced by the first derivative, (0.14-0.01)S, VT derivative of time component and 2V'' the second derivative. Sigma is 0.48.

    Are these maybe the "instructions"...? (If "substitute S in formula dv with the inverse of e(2/5)" indicates "the question", then I'm not understanding the question.)

    Okay, the above formula (VS(a-d).... is now filled in with the numbers and derivatives. However, since S is in the formula it is still a function for the stock price. Thus, in order to value the derivative (call option e.g.) S has to be substituted by V.
    V in this case is the inverse of V = S(1/1)e0.4(1-t) .

    Lacking a context (additive inverse, multiplicative inverse, functional inverse if we're saying "y = e^x; find inverse statement, given x = 2/5", etc), I'm not sure.


    No idea. See above.


    What is the "ito process"? Does "Vs" mean "the first derivative of V with respect to s, or with respect to S, or with respect [see below] to [tex]\delta[/tex])? (Similar question for "Vss".)
    Vs means the first derivative of the formula S(1/1)e0.4(1-t) . This applies to Vss (second derivative) and Vt (derivative of time) as well

    As originally posted, there was no [tex]\alpha,\, \delta,\,[/tex] or [tex]\sigma.[/tex] How do these relate?

    Also, does the above mean that the "ito lemma" is as follows?

    (VS(αd)S+Vt+12V(σS)2)Dt+VσSd

    yes, see above. This is the formula that has to filled in.

    Which derivative is this? What do you mean by "the value" of the derivative, when the right-hand side does not actually simplify to a numerical value? What do you mean by "can be described by"?

    You have to take the derivatives of equation S(1/1)e0.4(1-t) fill it into the equation (ito lemma) and substitute S by the inverse of S(1/1)e0.4(1-t) . in order to value a derivative (a call option or contract that depends on the value of a stock)

    How did you arrive at this?

    Taking the derivatives of the formula V = S(1/1)e0.4(1-t) and fill it into the equation. However, It think my derivatives of the equation are wrong.
    Which formula? How?

    Please reply showing all of your work. Thank you!

  5. #5
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    Quote Originally Posted by nasdreas View Post
    Dear all,

    I'm running into the following problem and I would like to have some assistance since I do not understand how to inverse a calculation and substitute it into an equation.

    Assume the following: V = S(1/1)e0.4(1-0) --> e(2/5)S and the the first derivative is e(2/5)
    Yes [tex]v = s^{(1/1)} * e^{\{0.4(1-0)\}} = e^{(2/5)}s \implies \dfrac{dv}{ds} = e^{(2/5)}.[/tex]

    If that was a question.

    Now I have to fill in an equation with the first derivative which looks like:
    dv = (e(2/5)*(0.14-0.01)S)
    This is nonsense.

    [tex]dv = e^{(2/5)} \ ds \ne e^{(2/5)} * (0.14 - 0.01) S.[/tex]

    You may be thinking that

    [tex]\Delta v \approx e{(2/5)} * \Delta s[/tex], which is true.

    Then the question is to substitute S in formula dv with the inverse of e(2/5)
    This makes no sense because S is not in an equation for dv.

    Can someone explain what the inverse of e(2/5) is? I thought LN(2/5)?
    As stapel explained, there are many kinds of inverse. Which type are you talking about. By the way

    [tex]y = e^{(2/5)} * \Delta s \implies ln(y) = \dfrac{2}{5} + ln \left ( \Delta s \right ).[/tex]

    if that was what you are getting at.

    I suspect you are copying things inaccurately because you do not know what they mean.

    And you have not provided a definition or a link to explain what the Ito process is.

  6. #6
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    Are we talking about this? https://en.wikipedia.org/wiki/It%C3%B4%27s_lemma

    If so, that seems quite an advanced thing to tackle for someone who...doesn't know...calculus...

    EDIT: Also, I love how we are using "derivative" in the finance sense ("valuing a derivative"), ("for instance a call option or put option or any other contract that derives its value from the stock"), and "derivative" in the calculus sense in the same thread

    All with no attempt made to...differentiate...the two terms
    Last edited by j-astron; 03-05-2018 at 12:06 AM.

  7. #7
    Quote Originally Posted by j-astron View Post
    Are we talking about this? https://en.wikipedia.org/wiki/It%C3%B4%27s_lemma

    If so, that seems quite an advanced thing to tackle for someone who...doesn't know...calculus...

    EDIT: Also, I love how we are using "derivative" in the finance sense ("valuing a derivative"), ("for instance a call option or put option or any other contract that derives its value from the stock"), and "derivative" in the calculus sense in the same thread

    All with no attempt made to...differentiate...the two terms

    Yes we are talking about that thing haha.
    With respect to the latter, that is why I put call option between the brackets haha thought that it was confusing to.

    Well if someones knows the first and second derivative of V = S(1/1)e0.4(1-t)
    and the derivative with respect to time (t) then it would already be helpful.
    (Derivative --> no option in this regard

  8. #8
    Someone that knows how to derive the derivatives?

  9. #9
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    Quote Originally Posted by nasdreas View Post
    Yes we are talking about that thing haha.
    With respect to the latter, that is why I put call option between the brackets haha thought that it was confusing to.

    Well if someones knows the first and second derivative of V = S(1/1)e0.4(1-t)
    and the derivative with respect to time (t) then it would already be helpful.
    (Derivative --> no option in this regard
    First of all, 1/1 = 1, and anything to the power of 1 is just equal to the thing itself. So I'm not sure why you keep writing S^(1/1) when you can just write S.

    So you want to know

    [tex] \dfrac{dV}{dt} = \dfrac{d}{dt}Se^{0.4(1-t)} [/tex]

    My question would be: does S depend on t, or is it a constant? The answer will be very different in either case.

    Here is some information that may be of use regardless:
    https://en.wikipedia.org/wiki/Expone...tial_equations

  10. #10
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    inverse

    Quote Originally Posted by nasdreas View Post
    I think it helps if I clarify the context of the question.
    The question is about valuing a derivative using ito process and ito lemma.

    For instance, alpha (a) is 0.14, delta (s) is 0.01 and sigma (o) is 0.48

    Formula ito lemma: (VS(a-d)S + Vt + 1/2Vss(oS)2 ) Dt + VsoSdz

    Vs = first derivative, Vss is second derivative Vt is first derivative with respect to time.

    The value of the derivative can be described by: V = S(1/1)e0.4*(1-t)
    I think that Vs = e0.4s and the second derivative is 0.

    Then the formula becomes:
    (e0.4S(0.14-0.1)S + Vt(???) + 1/2*)*(0.48S)2 ) Dt + e0.4s*0.48S dz
    (e0.4S(0.14-0.1)S + Vt(???))Dt + e0.4s*0.48S dz

    Questions:
    1. I don't know what the inverse is;
    in general, an inverse is interchanging independent and dependent variables.
    example v= 4t, then t=(1/4)v
    one equation is inverse of the other.
    another example;
    1/2=sinx, then sin-1(1/2)=x.

    another;
    y=x2; dy/dx=2x
    x=y1/2 dx/dy=1/2y-1/2
    dy/dx is the inverse of dx/dy
    Last edited by sinx; 03-14-2018 at 10:58 PM.

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