Pricing under risk-neutral probabilities for weird derivatives?
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I would really appreciate some help to value a weird derivative that I've found in an assignment:
$$ X=(S_{T_1}-k)^{+} = max(S_{T_{1}}-k;0) $$
which expires at time $T_{2}$ and uses the price at time $T_{1}$ (therefore $t<T_1<T_2$), using "R" (risk-neutral) probabilities. I tried to solve by doing:
$$ V_t=S_t times E_R [(S_{T_1}-k)times1_{(S_{T_1}>k)}times S_{T_2}^{-1} | F_t] $$
where $1_{(S_{T_2}>k)}$ is a function that takes a value of 1 if the condition is met and 0 if it's not, and $F_t$ is the information set at $t$. Solved it assuming $S_t=S_0times e^{(r+sigma^{2}/2)times t+sigmatimes W_t}$ where $W_t$ is a Brownian Motion process, and got the expression:
$$ V_t=S_t times N(d_1) - ktimes N(d_2) $$
where $d_1=frac{ln(K)+(r+frac{sigma2}{2})times(T_{2}-T_{1})}{sigma times sqrt{T_{2}-T_{1}}}$
and $d_2=frac{ln(K)+(r+frac{sigma2}{2})times(T_{2}-t)}{sigma times sqrt{T_{2}-t}}$ but I'm not sure this is even close to being correct.
Then I'm asked to price the same derivative under $Q$ (risk neutral probabilities) given that $T_1<t<T_2$.
Thanks in advance to whoever can provide some assistance.
option-pricing black-scholes derivatives valuation
edited May 21 at 13:27
bhutes
71810 bronze badges
asked May 19 at 11:25
BorisDBorisD
163 bronze badges
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|
show 7 more comments
$begingroup$
I would really appreciate some help to value a weird derivative that I've found in an assignment:
$$ X=(S_{T_1}-k)^{+} = max(S_{T_{1}}-k;0) $$
which expires at time $T_{2}$ and uses the price at time $T_{1}$ (therefore $t<T_1<T_2$), using "R" (risk-neutral) probabilities. I tried to solve by doing:
$$ V_t=S_t times E_R [(S_{T_1}-k)times1_{(S_{T_1}>k)}times S_{T_2}^{-1} | F_t] $$
where $1_{(S_{T_2}>k)}$ is a function that takes a value of 1 if the condition is met and 0 if it's not, and $F_t$ is the information set at $t$. Solved it assuming $S_t=S_0times e^{(r+sigma^{2}/2)times t+sigmatimes W_t}$ where $W_t$ is a Brownian Motion process, and got the expression:
$$ V_t=S_t times N(d_1) - ktimes N(d_2) $$
where $d_1=frac{ln(K)+(r+frac{sigma2}{2})times(T_{2}-T_{1})}{sigma times sqrt{T_{2}-T_{1}}}$
and $d_2=frac{ln(K)+(r+frac{sigma2}{2})times(T_{2}-t)}{sigma times sqrt{T_{2}-t}}$ but I'm not sure this is even close to being correct.
Then I'm asked to price the same derivative under $Q$ (risk neutral probabilities) given that $T_1<t<T_2$.
Thanks in advance to whoever can provide some assistance.
option-pricing black-scholes derivatives valuation
edited May 21 at 13:27
bhutes
71810 bronze badges
asked May 19 at 11:25
BorisDBorisD
163 bronze badges
$endgroup$
$begingroup$
What is going on with first line (double equal sign?) and 3rd line (how did you just factor $S_t$ but get $S_{T_2}^{-1}$ inside?). Use 1_{} for indicators
$endgroup$
– Makina
May 19 at 11:54
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Fixed line 1. Thanks! In the third line, instead of using bonds to get the neutral-risk probabilities in order to calculate the replicating portfolio ($V_t$), I use an asset (in this case, the underlying asset). This is: $$ Vt=S_t * E_{R} [X * S_{T} | F_{t}] $$ Where $X$ is the function of the derivative, $t$ is any time of valuation prior to the expiration date, and $T$ is the time of expiration.
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– BorisD
May 19 at 12:40
3
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$Eleft(e^{-rT_2}(S_{T_1}-K)^+right) = e^{-r(T_2-T_1)}Eleft(e^{-rT_1}(S_{T_1}-K)^+right)$.
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– Gordon
May 19 at 12:41
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This is still unclear. The payoff formula does not use S(T2), as claimed in the text.
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– dm63
May 19 at 13:19
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Sorry. Fixed. Payoff formula does not use $S_{T_2}$.
$endgroup$
– BorisD
May 19 at 13:21
|
show 7 more comments
$begingroup$
I would really appreciate some help to value a weird derivative that I've found in an assignment:
$$ X=(S_{T_1}-k)^{+} = max(S_{T_{1}}-k;0) $$
which expires at time $T_{2}$ and uses the price at time $T_{1}$ (therefore $t<T_1<T_2$), using "R" (risk-neutral) probabilities. I tried to solve by doing:
$$ V_t=S_t times E_R [(S_{T_1}-k)times1_{(S_{T_1}>k)}times S_{T_2}^{-1} | F_t] $$
where $1_{(S_{T_2}>k)}$ is a function that takes a value of 1 if the condition is met and 0 if it's not, and $F_t$ is the information set at $t$. Solved it assuming $S_t=S_0times e^{(r+sigma^{2}/2)times t+sigmatimes W_t}$ where $W_t$ is a Brownian Motion process, and got the expression:
$$ V_t=S_t times N(d_1) - ktimes N(d_2) $$
where $d_1=frac{ln(K)+(r+frac{sigma2}{2})times(T_{2}-T_{1})}{sigma times sqrt{T_{2}-T_{1}}}$
and $d_2=frac{ln(K)+(r+frac{sigma2}{2})times(T_{2}-t)}{sigma times sqrt{T_{2}-t}}$ but I'm not sure this is even close to being correct.
Then I'm asked to price the same derivative under $Q$ (risk neutral probabilities) given that $T_1<t<T_2$.
Thanks in advance to whoever can provide some assistance.
option-pricing black-scholes derivatives valuation
edited May 21 at 13:27
bhutes
71810 bronze badges
asked May 19 at 11:25
BorisDBorisD
163 bronze badges
$endgroup$
I would really appreciate some help to value a weird derivative that I've found in an assignment:
$$ X=(S_{T_1}-k)^{+} = max(S_{T_{1}}-k;0) $$
which expires at time $T_{2}$ and uses the price at time $T_{1}$ (therefore $t<T_1<T_2$), using "R" (risk-neutral) probabilities. I tried to solve by doing:
$$ V_t=S_t times E_R [(S_{T_1}-k)times1_{(S_{T_1}>k)}times S_{T_2}^{-1} | F_t] $$
where $1_{(S_{T_2}>k)}$ is a function that takes a value of 1 if the condition is met and 0 if it's not, and $F_t$ is the information set at $t$. Solved it assuming $S_t=S_0times e^{(r+sigma^{2}/2)times t+sigmatimes W_t}$ where $W_t$ is a Brownian Motion process, and got the expression:
$$ V_t=S_t times N(d_1) - ktimes N(d_2) $$
where $d_1=frac{ln(K)+(r+frac{sigma2}{2})times(T_{2}-T_{1})}{sigma times sqrt{T_{2}-T_{1}}}$
and $d_2=frac{ln(K)+(r+frac{sigma2}{2})times(T_{2}-t)}{sigma times sqrt{T_{2}-t}}$ but I'm not sure this is even close to being correct.
Then I'm asked to price the same derivative under $Q$ (risk neutral probabilities) given that $T_1<t<T_2$.
Thanks in advance to whoever can provide some assistance.
option-pricing black-scholes derivatives valuation
option-pricing black-scholes derivatives valuation
edited May 21 at 13:27
bhutes
71810 bronze badges
asked May 19 at 11:25
BorisDBorisD
163 bronze badges
edited May 21 at 13:27
bhutes
71810 bronze badges
asked May 19 at 11:25
BorisDBorisD
163 bronze badges
edited May 21 at 13:27
bhutes
71810 bronze badges
edited May 21 at 13:27
bhutes
71810 bronze badges
edited May 21 at 13:27
bhutes
71810 bronze badges
71810 bronze badges
asked May 19 at 11:25
BorisDBorisD
163 bronze badges
asked May 19 at 11:25
BorisDBorisD
163 bronze badges
asked May 19 at 11:25
BorisDBorisD
163 bronze badges
163 bronze badges
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What is going on with first line (double equal sign?) and 3rd line (how did you just factor $S_t$ but get $S_{T_2}^{-1}$ inside?). Use 1_{} for indicators
$endgroup$
– Makina
May 19 at 11:54
$begingroup$
Fixed line 1. Thanks! In the third line, instead of using bonds to get the neutral-risk probabilities in order to calculate the replicating portfolio ($V_t$), I use an asset (in this case, the underlying asset). This is: $$ Vt=S_t * E_{R} [X * S_{T} | F_{t}] $$ Where $X$ is the function of the derivative, $t$ is any time of valuation prior to the expiration date, and $T$ is the time of expiration.
$endgroup$
– BorisD
May 19 at 12:40
3
$begingroup$
$Eleft(e^{-rT_2}(S_{T_1}-K)^+right) = e^{-r(T_2-T_1)}Eleft(e^{-rT_1}(S_{T_1}-K)^+right)$.
$endgroup$
– Gordon
May 19 at 12:41
$begingroup$
This is still unclear. The payoff formula does not use S(T2), as claimed in the text.
$endgroup$
– dm63
May 19 at 13:19
$begingroup$
Sorry. Fixed. Payoff formula does not use $S_{T_2}$.
$endgroup$
– BorisD
May 19 at 13:21
|
show 7 more comments
$begingroup$
What is going on with first line (double equal sign?) and 3rd line (how did you just factor $S_t$ but get $S_{T_2}^{-1}$ inside?). Use 1_{} for indicators
$endgroup$
– Makina
May 19 at 11:54
$begingroup$
Fixed line 1. Thanks! In the third line, instead of using bonds to get the neutral-risk probabilities in order to calculate the replicating portfolio ($V_t$), I use an asset (in this case, the underlying asset). This is: $$ Vt=S_t * E_{R} [X * S_{T} | F_{t}] $$ Where $X$ is the function of the derivative, $t$ is any time of valuation prior to the expiration date, and $T$ is the time of expiration.
$endgroup$
– BorisD
May 19 at 12:40
3
$begingroup$
$Eleft(e^{-rT_2}(S_{T_1}-K)^+right) = e^{-r(T_2-T_1)}Eleft(e^{-rT_1}(S_{T_1}-K)^+right)$.
$endgroup$
– Gordon
May 19 at 12:41
$begingroup$
This is still unclear. The payoff formula does not use S(T2), as claimed in the text.
$endgroup$
– dm63
May 19 at 13:19
$begingroup$
Sorry. Fixed. Payoff formula does not use $S_{T_2}$.
$endgroup$
– BorisD
May 19 at 13:21
$begingroup$
What is going on with first line (double equal sign?) and 3rd line (how did you just factor $S_t$ but get $S_{T_2}^{-1}$ inside?). Use 1_{} for indicators
$endgroup$
– Makina
May 19 at 11:54
$begingroup$
What is going on with first line (double equal sign?) and 3rd line (how did you just factor $S_t$ but get $S_{T_2}^{-1}$ inside?). Use 1_{} for indicators
$endgroup$
– Makina
May 19 at 11:54
$begingroup$
Fixed line 1. Thanks! In the third line, instead of using bonds to get the neutral-risk probabilities in order to calculate the replicating portfolio ($V_t$), I use an asset (in this case, the underlying asset). This is: $$ Vt=S_t * E_{R} [X * S_{T} | F_{t}] $$ Where $X$ is the function of the derivative, $t$ is any time of valuation prior to the expiration date, and $T$ is the time of expiration.
$endgroup$
– BorisD
May 19 at 12:40
$begingroup$
Fixed line 1. Thanks! In the third line, instead of using bonds to get the neutral-risk probabilities in order to calculate the replicating portfolio ($V_t$), I use an asset (in this case, the underlying asset). This is: $$ Vt=S_t * E_{R} [X * S_{T} | F_{t}] $$ Where $X$ is the function of the derivative, $t$ is any time of valuation prior to the expiration date, and $T$ is the time of expiration.
$endgroup$
– BorisD
May 19 at 12:40
3
3
$begingroup$
$Eleft(e^{-rT_2}(S_{T_1}-K)^+right) = e^{-r(T_2-T_1)}Eleft(e^{-rT_1}(S_{T_1}-K)^+right)$.
$endgroup$
– Gordon
May 19 at 12:41
$begingroup$
$Eleft(e^{-rT_2}(S_{T_1}-K)^+right) = e^{-r(T_2-T_1)}Eleft(e^{-rT_1}(S_{T_1}-K)^+right)$.
$endgroup$
– Gordon
May 19 at 12:41
$begingroup$
This is still unclear. The payoff formula does not use S(T2), as claimed in the text.
$endgroup$
– dm63
May 19 at 13:19
$begingroup$
This is still unclear. The payoff formula does not use S(T2), as claimed in the text.
$endgroup$
– dm63
May 19 at 13:19
$begingroup$
Sorry. Fixed. Payoff formula does not use $S_{T_2}$.
$endgroup$
– BorisD
May 19 at 13:21
$begingroup$
Sorry. Fixed. Payoff formula does not use $S_{T_2}$.
$endgroup$
– BorisD
May 19 at 13:21
|
show 7 more comments
1 Answer
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@Gordon has already given the answer but here is a little more notes to it...
At time time $T_2$ the holder receives $X=(S_{T_1}-K)^+$. According to Risk Neutral Valuation the value at time $t$ $(t<T_1<T_2)$ is $$V_t = e^{-r(T_2-t)}E_t[(S_{T_1}-K)^+] = \
e^{-r(T_2-t+T_1-T_1)}E_t[(S_{T_1}-K)^+]=\
e^{-r(T_2-T_1)}e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]
$$
$e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]$ is the value of a Call Option at time $t$ with expiration at time $T_1$. This is simply given by the Black-Sholes formula so $e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]=C_{BS}(S_t,t;T_1)$
$$
V_t=e^{-r(T_2-T_1)}e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]=e^{-r(T_2-T_1)}C_{BS}(S_t,t;T_1)
$$
For $T_1<t<T_2$ then $(S_{T_1} - K )^+$ is measurable so $E_t[(S_{T_1} - K )^+]=(S_{T_1} - K )^+$. This means you know exactly what you get and only have to discount the pay-off: $V_t=e^{-r(T_2-1)}(S_{T_1} - K )^+$
answered May 19 at 14:07
SanjaySanjay
9724 silver badges17 bronze badges
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1
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You've missed a minus sign in front of $r$, which you can also see by the fact that if $r>0$ then by your formula $V_t>C_{BS}(S_t,t;T_1)$ which is inconsistent with positive rates as it's always preferable to receive a payment at $T_1$ than at $T_2>T_1$ (as you can earn the rate $r$ between $T_1$ and $T_2$ if you've been paid at $T_1<T_2$).
$endgroup$
– Daneel Olivaw
May 19 at 14:15
add a comment |
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@Gordon has already given the answer but here is a little more notes to it...
At time time $T_2$ the holder receives $X=(S_{T_1}-K)^+$. According to Risk Neutral Valuation the value at time $t$ $(t<T_1<T_2)$ is $$V_t = e^{-r(T_2-t)}E_t[(S_{T_1}-K)^+] = \
e^{-r(T_2-t+T_1-T_1)}E_t[(S_{T_1}-K)^+]=\
e^{-r(T_2-T_1)}e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]
$$
$e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]$ is the value of a Call Option at time $t$ with expiration at time $T_1$. This is simply given by the Black-Sholes formula so $e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]=C_{BS}(S_t,t;T_1)$
$$
V_t=e^{-r(T_2-T_1)}e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]=e^{-r(T_2-T_1)}C_{BS}(S_t,t;T_1)
$$
For $T_1<t<T_2$ then $(S_{T_1} - K )^+$ is measurable so $E_t[(S_{T_1} - K )^+]=(S_{T_1} - K )^+$. This means you know exactly what you get and only have to discount the pay-off: $V_t=e^{-r(T_2-1)}(S_{T_1} - K )^+$
answered May 19 at 14:07
SanjaySanjay
9724 silver badges17 bronze badges
$endgroup$
1
$begingroup$
You've missed a minus sign in front of $r$, which you can also see by the fact that if $r>0$ then by your formula $V_t>C_{BS}(S_t,t;T_1)$ which is inconsistent with positive rates as it's always preferable to receive a payment at $T_1$ than at $T_2>T_1$ (as you can earn the rate $r$ between $T_1$ and $T_2$ if you've been paid at $T_1<T_2$).
$endgroup$
– Daneel Olivaw
May 19 at 14:15
add a comment |
$begingroup$
@Gordon has already given the answer but here is a little more notes to it...
At time time $T_2$ the holder receives $X=(S_{T_1}-K)^+$. According to Risk Neutral Valuation the value at time $t$ $(t<T_1<T_2)$ is $$V_t = e^{-r(T_2-t)}E_t[(S_{T_1}-K)^+] = \
e^{-r(T_2-t+T_1-T_1)}E_t[(S_{T_1}-K)^+]=\
e^{-r(T_2-T_1)}e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]
$$
$e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]$ is the value of a Call Option at time $t$ with expiration at time $T_1$. This is simply given by the Black-Sholes formula so $e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]=C_{BS}(S_t,t;T_1)$
$$
V_t=e^{-r(T_2-T_1)}e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]=e^{-r(T_2-T_1)}C_{BS}(S_t,t;T_1)
$$
For $T_1<t<T_2$ then $(S_{T_1} - K )^+$ is measurable so $E_t[(S_{T_1} - K )^+]=(S_{T_1} - K )^+$. This means you know exactly what you get and only have to discount the pay-off: $V_t=e^{-r(T_2-1)}(S_{T_1} - K )^+$
answered May 19 at 14:07
SanjaySanjay
9724 silver badges17 bronze badges
$endgroup$
1
$begingroup$
You've missed a minus sign in front of $r$, which you can also see by the fact that if $r>0$ then by your formula $V_t>C_{BS}(S_t,t;T_1)$ which is inconsistent with positive rates as it's always preferable to receive a payment at $T_1$ than at $T_2>T_1$ (as you can earn the rate $r$ between $T_1$ and $T_2$ if you've been paid at $T_1<T_2$).
$endgroup$
– Daneel Olivaw
May 19 at 14:15
add a comment |
$begingroup$
@Gordon has already given the answer but here is a little more notes to it...
At time time $T_2$ the holder receives $X=(S_{T_1}-K)^+$. According to Risk Neutral Valuation the value at time $t$ $(t<T_1<T_2)$ is $$V_t = e^{-r(T_2-t)}E_t[(S_{T_1}-K)^+] = \
e^{-r(T_2-t+T_1-T_1)}E_t[(S_{T_1}-K)^+]=\
e^{-r(T_2-T_1)}e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]
$$
$e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]$ is the value of a Call Option at time $t$ with expiration at time $T_1$. This is simply given by the Black-Sholes formula so $e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]=C_{BS}(S_t,t;T_1)$
$$
V_t=e^{-r(T_2-T_1)}e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]=e^{-r(T_2-T_1)}C_{BS}(S_t,t;T_1)
$$
For $T_1<t<T_2$ then $(S_{T_1} - K )^+$ is measurable so $E_t[(S_{T_1} - K )^+]=(S_{T_1} - K )^+$. This means you know exactly what you get and only have to discount the pay-off: $V_t=e^{-r(T_2-1)}(S_{T_1} - K )^+$
answered May 19 at 14:07
SanjaySanjay
9724 silver badges17 bronze badges
$endgroup$
@Gordon has already given the answer but here is a little more notes to it...
At time time $T_2$ the holder receives $X=(S_{T_1}-K)^+$. According to Risk Neutral Valuation the value at time $t$ $(t<T_1<T_2)$ is $$V_t = e^{-r(T_2-t)}E_t[(S_{T_1}-K)^+] = \
e^{-r(T_2-t+T_1-T_1)}E_t[(S_{T_1}-K)^+]=\
e^{-r(T_2-T_1)}e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]
$$
$e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]$ is the value of a Call Option at time $t$ with expiration at time $T_1$. This is simply given by the Black-Sholes formula so $e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]=C_{BS}(S_t,t;T_1)$
$$
V_t=e^{-r(T_2-T_1)}e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]=e^{-r(T_2-T_1)}C_{BS}(S_t,t;T_1)
$$
For $T_1<t<T_2$ then $(S_{T_1} - K )^+$ is measurable so $E_t[(S_{T_1} - K )^+]=(S_{T_1} - K )^+$. This means you know exactly what you get and only have to discount the pay-off: $V_t=e^{-r(T_2-1)}(S_{T_1} - K )^+$
answered May 19 at 14:07
SanjaySanjay
9724 silver badges17 bronze badges
edited May 19 at 14:23
answered May 19 at 14:07
SanjaySanjay
9724 silver badges17 bronze badges
answered May 19 at 14:07
SanjaySanjay
9724 silver badges17 bronze badges
answered May 19 at 14:07
SanjaySanjay
9724 silver badges17 bronze badges
9724 silver badges17 bronze badges
1
$begingroup$
You've missed a minus sign in front of $r$, which you can also see by the fact that if $r>0$ then by your formula $V_t>C_{BS}(S_t,t;T_1)$ which is inconsistent with positive rates as it's always preferable to receive a payment at $T_1$ than at $T_2>T_1$ (as you can earn the rate $r$ between $T_1$ and $T_2$ if you've been paid at $T_1<T_2$).
$endgroup$
– Daneel Olivaw
May 19 at 14:15
add a comment |
1
$begingroup$
You've missed a minus sign in front of $r$, which you can also see by the fact that if $r>0$ then by your formula $V_t>C_{BS}(S_t,t;T_1)$ which is inconsistent with positive rates as it's always preferable to receive a payment at $T_1$ than at $T_2>T_1$ (as you can earn the rate $r$ between $T_1$ and $T_2$ if you've been paid at $T_1<T_2$).
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– Daneel Olivaw
May 19 at 14:15
1
1
$begingroup$
You've missed a minus sign in front of $r$, which you can also see by the fact that if $r>0$ then by your formula $V_t>C_{BS}(S_t,t;T_1)$ which is inconsistent with positive rates as it's always preferable to receive a payment at $T_1$ than at $T_2>T_1$ (as you can earn the rate $r$ between $T_1$ and $T_2$ if you've been paid at $T_1<T_2$).
$endgroup$
– Daneel Olivaw
May 19 at 14:15
$begingroup$
You've missed a minus sign in front of $r$, which you can also see by the fact that if $r>0$ then by your formula $V_t>C_{BS}(S_t,t;T_1)$ which is inconsistent with positive rates as it's always preferable to receive a payment at $T_1$ than at $T_2>T_1$ (as you can earn the rate $r$ between $T_1$ and $T_2$ if you've been paid at $T_1<T_2$).
$endgroup$
– Daneel Olivaw
May 19 at 14:15
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$begingroup$
What is going on with first line (double equal sign?) and 3rd line (how did you just factor $S_t$ but get $S_{T_2}^{-1}$ inside?). Use 1_{} for indicators
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– Makina
May 19 at 11:54
$begingroup$
Fixed line 1. Thanks! In the third line, instead of using bonds to get the neutral-risk probabilities in order to calculate the replicating portfolio ($V_t$), I use an asset (in this case, the underlying asset). This is: $$ Vt=S_t * E_{R} [X * S_{T} | F_{t}] $$ Where $X$ is the function of the derivative, $t$ is any time of valuation prior to the expiration date, and $T$ is the time of expiration.
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– BorisD
May 19 at 12:40
3
$begingroup$
$Eleft(e^{-rT_2}(S_{T_1}-K)^+right) = e^{-r(T_2-T_1)}Eleft(e^{-rT_1}(S_{T_1}-K)^+right)$.
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– Gordon
May 19 at 12:41
$begingroup$
This is still unclear. The payoff formula does not use S(T2), as claimed in the text.
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– dm63
May 19 at 13:19
$begingroup$
Sorry. Fixed. Payoff formula does not use $S_{T_2}$.
$endgroup$
– BorisD
May 19 at 13:21