Problem at deriving Bachelier formula with interest ratesWhy is the black-scholes model arbitrage free when σ>0?Square of arithmetic brownian motion processStochastic process and brownian motionpricing option with two stocksChange-of-measure: Dynamics of $log(S_t)$ with $S_t$ as numeraireBlack & Scholes with stochastic interest ratePricing caplet with Bachelier (normal dynamic) using forward measureQuestion about Stochastic Calculus,(change of measure)?Ultra Powerfull Vibrato Montecarlo for delta sensitivities of a not regular payoffEquivalence of Put Pricing Formulas
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Problem at deriving Bachelier formula with interest rates
Why is the black-scholes model arbitrage free when σ>0?Square of arithmetic brownian motion processStochastic process and brownian motionpricing option with two stocksChange-of-measure: Dynamics of $log(S_t)$ with $S_t$ as numeraireBlack & Scholes with stochastic interest ratePricing caplet with Bachelier (normal dynamic) using forward measureQuestion about Stochastic Calculus,(change of measure)?Ultra Powerfull Vibrato Montecarlo for delta sensitivities of a not regular payoffEquivalence of Put Pricing Formulas
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
In the Bachelier model, I have difficulties with a certain step. I want to figure out the distribution of $S_T$, which is the price process in the Bachelier model.
So far I could state that ($mathbbQ$ is the EMM):
begineqnarray
dS_t = r S_t dt + sigma W^mathbbQ_t labelSDE2
endeqnarray
and with that
begineqnarray
S_T = S_0 e^rT + intlimits_0^Tsigma e^r(T-s) dW^mathbbQ_s
endeqnarray
Now I have found a book that states that $S_T$ has distribution:
begineqnarray
S_T sim mathscr N left(S_0 e^rT, sqrtfracsigma^2-sigma^2e^-2rT2r right)
endeqnarray
I do not understand why this should be, maybe my skills in stochastic integration are not sufficient.
Thank you for taking your time!
options stochastic-calculus pricing models
asked Jun 26 at 7:43
cruiser0223cruiser0223
184 bronze badges
$endgroup$
add a comment |
$begingroup$
In the Bachelier model, I have difficulties with a certain step. I want to figure out the distribution of $S_T$, which is the price process in the Bachelier model.
So far I could state that ($mathbbQ$ is the EMM):
begineqnarray
dS_t = r S_t dt + sigma W^mathbbQ_t labelSDE2
endeqnarray
and with that
begineqnarray
S_T = S_0 e^rT + intlimits_0^Tsigma e^r(T-s) dW^mathbbQ_s
endeqnarray
Now I have found a book that states that $S_T$ has distribution:
begineqnarray
S_T sim mathscr N left(S_0 e^rT, sqrtfracsigma^2-sigma^2e^-2rT2r right)
endeqnarray
I do not understand why this should be, maybe my skills in stochastic integration are not sufficient.
Thank you for taking your time!
options stochastic-calculus pricing models
asked Jun 26 at 7:43
cruiser0223cruiser0223
184 bronze badges
$endgroup$
4
$begingroup$
This is a direct application of Ito's isometry: en.m.wikipedia.org/wiki/It%C3%B4_isometry it gives you the mean (= 0) and variance ($= int f^2(u)du$) of a Wiener integral $int f(u) dW(u)$.
$endgroup$
– byouness
Jun 26 at 8:00
add a comment |
$begingroup$
In the Bachelier model, I have difficulties with a certain step. I want to figure out the distribution of $S_T$, which is the price process in the Bachelier model.
So far I could state that ($mathbbQ$ is the EMM):
begineqnarray
dS_t = r S_t dt + sigma W^mathbbQ_t labelSDE2
endeqnarray
and with that
begineqnarray
S_T = S_0 e^rT + intlimits_0^Tsigma e^r(T-s) dW^mathbbQ_s
endeqnarray
Now I have found a book that states that $S_T$ has distribution:
begineqnarray
S_T sim mathscr N left(S_0 e^rT, sqrtfracsigma^2-sigma^2e^-2rT2r right)
endeqnarray
I do not understand why this should be, maybe my skills in stochastic integration are not sufficient.
Thank you for taking your time!
options stochastic-calculus pricing models
asked Jun 26 at 7:43
cruiser0223cruiser0223
184 bronze badges
$endgroup$
In the Bachelier model, I have difficulties with a certain step. I want to figure out the distribution of $S_T$, which is the price process in the Bachelier model.
So far I could state that ($mathbbQ$ is the EMM):
begineqnarray
dS_t = r S_t dt + sigma W^mathbbQ_t labelSDE2
endeqnarray
and with that
begineqnarray
S_T = S_0 e^rT + intlimits_0^Tsigma e^r(T-s) dW^mathbbQ_s
endeqnarray
Now I have found a book that states that $S_T$ has distribution:
begineqnarray
S_T sim mathscr N left(S_0 e^rT, sqrtfracsigma^2-sigma^2e^-2rT2r right)
endeqnarray
I do not understand why this should be, maybe my skills in stochastic integration are not sufficient.
Thank you for taking your time!
options stochastic-calculus pricing models
options stochastic-calculus pricing models
asked Jun 26 at 7:43
cruiser0223cruiser0223
184 bronze badges
asked Jun 26 at 7:43
cruiser0223cruiser0223
184 bronze badges
edited Jun 26 at 7:48
cruiser0223
asked Jun 26 at 7:43
cruiser0223cruiser0223
184 bronze badges
asked Jun 26 at 7:43
cruiser0223cruiser0223
184 bronze badges
asked Jun 26 at 7:43
cruiser0223cruiser0223
184 bronze badges
184 bronze badges
4
$begingroup$
This is a direct application of Ito's isometry: en.m.wikipedia.org/wiki/It%C3%B4_isometry it gives you the mean (= 0) and variance ($= int f^2(u)du$) of a Wiener integral $int f(u) dW(u)$.
$endgroup$
– byouness
Jun 26 at 8:00
add a comment |
4
$begingroup$
This is a direct application of Ito's isometry: en.m.wikipedia.org/wiki/It%C3%B4_isometry it gives you the mean (= 0) and variance ($= int f^2(u)du$) of a Wiener integral $int f(u) dW(u)$.
$endgroup$
– byouness
Jun 26 at 8:00
4
4
$begingroup$
This is a direct application of Ito's isometry: en.m.wikipedia.org/wiki/It%C3%B4_isometry it gives you the mean (= 0) and variance ($= int f^2(u)du$) of a Wiener integral $int f(u) dW(u)$.
$endgroup$
– byouness
Jun 26 at 8:00
$begingroup$
This is a direct application of Ito's isometry: en.m.wikipedia.org/wiki/It%C3%B4_isometry it gives you the mean (= 0) and variance ($= int f^2(u)du$) of a Wiener integral $int f(u) dW(u)$.
$endgroup$
– byouness
Jun 26 at 8:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As explained by @byouness, using Itô's Isometry, we get:
$$beginalign
V(S_T)&=V^mathbbQleft(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)
\[9pt]
&=E^mathbbQleft(left(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)^2right)-underbraceE^mathbbQleft(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)_=int_0^Tsigma e^r(T-s) E^mathbbQ(dW^mathbbQ_s)=0^2
\[-9pt]
&=E^mathbbQleft(int_0^Tsigma^2 e^2r(T-s) dsright)
endalign$$
The remaining integral is deterministic, thus:
$$V(S_T)=sigma^2left[-frace^2r(T-s)2rright]_s=0^s=T=sigma^2left(frace^2rT-12rright)$$
Note that your result is correct up to a minus sign. This is probably because the Bachelier dynamics for the stock price are also known as an Ornstein–Uhlenbeck process, which is normally defined with a minus sign in the drift, i.e.:
$$dS_t = colorred-r S_t dt + sigma W^mathbbQ_t$$
in which case the volatility is given by the expression in your original post.
answered Jun 26 at 9:11
Daneel OlivawDaneel Olivaw
3,3431 gold badge8 silver badges31 bronze badges
$endgroup$
add a comment |
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$begingroup$
As explained by @byouness, using Itô's Isometry, we get:
$$beginalign
V(S_T)&=V^mathbbQleft(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)
\[9pt]
&=E^mathbbQleft(left(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)^2right)-underbraceE^mathbbQleft(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)_=int_0^Tsigma e^r(T-s) E^mathbbQ(dW^mathbbQ_s)=0^2
\[-9pt]
&=E^mathbbQleft(int_0^Tsigma^2 e^2r(T-s) dsright)
endalign$$
The remaining integral is deterministic, thus:
$$V(S_T)=sigma^2left[-frace^2r(T-s)2rright]_s=0^s=T=sigma^2left(frace^2rT-12rright)$$
Note that your result is correct up to a minus sign. This is probably because the Bachelier dynamics for the stock price are also known as an Ornstein–Uhlenbeck process, which is normally defined with a minus sign in the drift, i.e.:
$$dS_t = colorred-r S_t dt + sigma W^mathbbQ_t$$
in which case the volatility is given by the expression in your original post.
answered Jun 26 at 9:11
Daneel OlivawDaneel Olivaw
3,3431 gold badge8 silver badges31 bronze badges
$endgroup$
add a comment |
$begingroup$
As explained by @byouness, using Itô's Isometry, we get:
$$beginalign
V(S_T)&=V^mathbbQleft(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)
\[9pt]
&=E^mathbbQleft(left(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)^2right)-underbraceE^mathbbQleft(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)_=int_0^Tsigma e^r(T-s) E^mathbbQ(dW^mathbbQ_s)=0^2
\[-9pt]
&=E^mathbbQleft(int_0^Tsigma^2 e^2r(T-s) dsright)
endalign$$
The remaining integral is deterministic, thus:
$$V(S_T)=sigma^2left[-frace^2r(T-s)2rright]_s=0^s=T=sigma^2left(frace^2rT-12rright)$$
Note that your result is correct up to a minus sign. This is probably because the Bachelier dynamics for the stock price are also known as an Ornstein–Uhlenbeck process, which is normally defined with a minus sign in the drift, i.e.:
$$dS_t = colorred-r S_t dt + sigma W^mathbbQ_t$$
in which case the volatility is given by the expression in your original post.
answered Jun 26 at 9:11
Daneel OlivawDaneel Olivaw
3,3431 gold badge8 silver badges31 bronze badges
$endgroup$
add a comment |
$begingroup$
As explained by @byouness, using Itô's Isometry, we get:
$$beginalign
V(S_T)&=V^mathbbQleft(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)
\[9pt]
&=E^mathbbQleft(left(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)^2right)-underbraceE^mathbbQleft(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)_=int_0^Tsigma e^r(T-s) E^mathbbQ(dW^mathbbQ_s)=0^2
\[-9pt]
&=E^mathbbQleft(int_0^Tsigma^2 e^2r(T-s) dsright)
endalign$$
The remaining integral is deterministic, thus:
$$V(S_T)=sigma^2left[-frace^2r(T-s)2rright]_s=0^s=T=sigma^2left(frace^2rT-12rright)$$
Note that your result is correct up to a minus sign. This is probably because the Bachelier dynamics for the stock price are also known as an Ornstein–Uhlenbeck process, which is normally defined with a minus sign in the drift, i.e.:
$$dS_t = colorred-r S_t dt + sigma W^mathbbQ_t$$
in which case the volatility is given by the expression in your original post.
answered Jun 26 at 9:11
Daneel OlivawDaneel Olivaw
3,3431 gold badge8 silver badges31 bronze badges
$endgroup$
As explained by @byouness, using Itô's Isometry, we get:
$$beginalign
V(S_T)&=V^mathbbQleft(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)
\[9pt]
&=E^mathbbQleft(left(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)^2right)-underbraceE^mathbbQleft(int_0^Tsigma e^r(T-s) dW^mathbbQ_sright)_=int_0^Tsigma e^r(T-s) E^mathbbQ(dW^mathbbQ_s)=0^2
\[-9pt]
&=E^mathbbQleft(int_0^Tsigma^2 e^2r(T-s) dsright)
endalign$$
The remaining integral is deterministic, thus:
$$V(S_T)=sigma^2left[-frace^2r(T-s)2rright]_s=0^s=T=sigma^2left(frace^2rT-12rright)$$
Note that your result is correct up to a minus sign. This is probably because the Bachelier dynamics for the stock price are also known as an Ornstein–Uhlenbeck process, which is normally defined with a minus sign in the drift, i.e.:
$$dS_t = colorred-r S_t dt + sigma W^mathbbQ_t$$
in which case the volatility is given by the expression in your original post.
answered Jun 26 at 9:11
Daneel OlivawDaneel Olivaw
3,3431 gold badge8 silver badges31 bronze badges
answered Jun 26 at 9:11
Daneel OlivawDaneel Olivaw
3,3431 gold badge8 silver badges31 bronze badges
answered Jun 26 at 9:11
Daneel OlivawDaneel Olivaw
3,3431 gold badge8 silver badges31 bronze badges
answered Jun 26 at 9:11
Daneel OlivawDaneel Olivaw
3,3431 gold badge8 silver badges31 bronze badges
3,3431 gold badge8 silver badges31 bronze badges
add a comment |
add a comment |
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4
$begingroup$
This is a direct application of Ito's isometry: en.m.wikipedia.org/wiki/It%C3%B4_isometry it gives you the mean (= 0) and variance ($= int f^2(u)du$) of a Wiener integral $int f(u) dW(u)$.
$endgroup$
– byouness
Jun 26 at 8:00