Expectation in a stochastic differential equationWhat is Ito's lemma used for in quantitative finance?Question about the stochastic differential equation in the Merton modelComputation of ExpectationSquare of arithmetic brownian motion processBaxter & Rennie HJM: differentiating Ito integralSimple HJM model, differentiating the bond priceStochastic Leibniz ruleStochastic differential equation of a Brownian MotionHow to calculate the product of forward rates with different reset times using Ito's lemma?For an Ito Process, $dlnX neq fracdXX$ and $(dlnX)^2 = (fracdXX)^2$, but $dlnX neq pm fracdXX$
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Expectation in a stochastic differential equation
What is Ito's lemma used for in quantitative finance?Question about the stochastic differential equation in the Merton modelComputation of ExpectationSquare of arithmetic brownian motion processBaxter & Rennie HJM: differentiating Ito integralSimple HJM model, differentiating the bond priceStochastic Leibniz ruleStochastic differential equation of a Brownian MotionHow to calculate the product of forward rates with different reset times using Ito's lemma?For an Ito Process, $dlnX neq fracdXX$ and $(dlnX)^2 = (fracdXX)^2$, but $dlnX neq pm fracdXX$
$begingroup$
I'm new to stochastic calculus, I want to find the mean of $X_2$ with $X_t = exp(W_t)$, with $W_t$ a Wiener process.
I used Ito's Lemma is arrive at the SDE:
beginalign
d(X_t) = frac12X_t dt + X_t dW_t
endalign
But how can I get the mean of $X_2$?
itos-lemma sde
asked 2 days ago
VictorVictor
764
$endgroup$
add a comment |
$begingroup$
I'm new to stochastic calculus, I want to find the mean of $X_2$ with $X_t = exp(W_t)$, with $W_t$ a Wiener process.
I used Ito's Lemma is arrive at the SDE:
beginalign
d(X_t) = frac12X_t dt + X_t dW_t
endalign
But how can I get the mean of $X_2$?
itos-lemma sde
asked 2 days ago
VictorVictor
764
$endgroup$
add a comment |
$begingroup$
I'm new to stochastic calculus, I want to find the mean of $X_2$ with $X_t = exp(W_t)$, with $W_t$ a Wiener process.
I used Ito's Lemma is arrive at the SDE:
beginalign
d(X_t) = frac12X_t dt + X_t dW_t
endalign
But how can I get the mean of $X_2$?
itos-lemma sde
asked 2 days ago
VictorVictor
764
$endgroup$
I'm new to stochastic calculus, I want to find the mean of $X_2$ with $X_t = exp(W_t)$, with $W_t$ a Wiener process.
I used Ito's Lemma is arrive at the SDE:
beginalign
d(X_t) = frac12X_t dt + X_t dW_t
endalign
But how can I get the mean of $X_2$?
itos-lemma sde
itos-lemma sde
asked 2 days ago
VictorVictor
764
asked 2 days ago
VictorVictor
764
edited 2 days ago
Victor
asked 2 days ago
VictorVictor
764
asked 2 days ago
VictorVictor
764
asked 2 days ago
VictorVictor
764
764
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assuming you are talking about unconditional expectation, in general you have
$$
mathbbE[X_t] = mathbbE[e^W_t] = e^mathbbE[W_t] + frac12textVar(W_t)
$$
which yields
$$
mathbbE[X_t]= e^frac12 t
$$
Hence,
$$ mathbbE[X_2]= e $$
answered 2 days ago
RafaelCRafaelC
1463
New contributor
RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
$endgroup$
– Victor
2 days ago
1
$begingroup$
@Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
$endgroup$
– RafaelC
2 days ago
add a comment |
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1 Answer
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oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Assuming you are talking about unconditional expectation, in general you have
$$
mathbbE[X_t] = mathbbE[e^W_t] = e^mathbbE[W_t] + frac12textVar(W_t)
$$
which yields
$$
mathbbE[X_t]= e^frac12 t
$$
Hence,
$$ mathbbE[X_2]= e $$
answered 2 days ago
RafaelCRafaelC
1463
New contributor
RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
$endgroup$
– Victor
2 days ago
1
$begingroup$
@Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
$endgroup$
– RafaelC
2 days ago
add a comment |
$begingroup$
Assuming you are talking about unconditional expectation, in general you have
$$
mathbbE[X_t] = mathbbE[e^W_t] = e^mathbbE[W_t] + frac12textVar(W_t)
$$
which yields
$$
mathbbE[X_t]= e^frac12 t
$$
Hence,
$$ mathbbE[X_2]= e $$
answered 2 days ago
RafaelCRafaelC
1463
New contributor
RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
$endgroup$
– Victor
2 days ago
1
$begingroup$
@Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
$endgroup$
– RafaelC
2 days ago
add a comment |
$begingroup$
Assuming you are talking about unconditional expectation, in general you have
$$
mathbbE[X_t] = mathbbE[e^W_t] = e^mathbbE[W_t] + frac12textVar(W_t)
$$
which yields
$$
mathbbE[X_t]= e^frac12 t
$$
Hence,
$$ mathbbE[X_2]= e $$
answered 2 days ago
RafaelCRafaelC
1463
New contributor
RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Assuming you are talking about unconditional expectation, in general you have
$$
mathbbE[X_t] = mathbbE[e^W_t] = e^mathbbE[W_t] + frac12textVar(W_t)
$$
which yields
$$
mathbbE[X_t]= e^frac12 t
$$
Hence,
$$ mathbbE[X_2]= e $$
answered 2 days ago
RafaelCRafaelC
1463
New contributor
RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
RafaelCRafaelC
1463
New contributor
RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
RafaelCRafaelC
1463
answered 2 days ago
RafaelCRafaelC
1463
1463
New contributor
RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
$endgroup$
– Victor
2 days ago
1
$begingroup$
@Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
$endgroup$
– RafaelC
2 days ago
add a comment |
$begingroup$
I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
$endgroup$
– Victor
2 days ago
1
$begingroup$
@Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
$endgroup$
– RafaelC
2 days ago
$begingroup$
I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
$endgroup$
– Victor
2 days ago
$begingroup$
I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
$endgroup$
– Victor
2 days ago
1
1
$begingroup$
@Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
$endgroup$
– RafaelC
2 days ago
$begingroup$
@Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
$endgroup$
– RafaelC
2 days ago
add a comment |
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