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Expectation in a stochastic differential equation
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Expectation in a stochastic differential equation
The Next CEO of Stack OverflowWhat 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, $dln{X} neq frac{dX}{X}$ and $(dln{X})^2 = (frac{dX}{X})^2$, but $dln{X} neq pm frac{dX}{X}$
$begingroup$
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:
begin{align}
d(X_t) = frac{1}{2}X_t dt + X_t dW_t
end{align}
But how can I get the mean of $X_2$?
itos-lemma sde
asked 6 hours ago
VictorVictor
614
$endgroup$
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:
begin{align}
d(X_t) = frac{1}{2}X_t dt + X_t dW_t
end{align}
But how can I get the mean of $X_2$?
itos-lemma sde
asked 6 hours ago
VictorVictor
614
$endgroup$
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:
begin{align}
d(X_t) = frac{1}{2}X_t dt + X_t dW_t
end{align}
But how can I get the mean of $X_2$?
itos-lemma sde
asked 6 hours ago
VictorVictor
614
$endgroup$
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:
begin{align}
d(X_t) = frac{1}{2}X_t dt + X_t dW_t
end{align}
But how can I get the mean of $X_2$?
itos-lemma sde
itos-lemma sde
asked 6 hours ago
VictorVictor
614
asked 6 hours ago
VictorVictor
614
edited 5 hours ago
Victor
asked 6 hours ago
VictorVictor
614
asked 6 hours ago
VictorVictor
614
asked 6 hours ago
VictorVictor
614
614
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Assuming you are talking about unconditional expectation, in general you have
$$
mathbb{E}[X_t] = mathbb{E}[e^{W_t}] = e^{mathbb{E}[W_t] + frac{1}{2}text{Var}(W_t) }
$$
which yields
$$
mathbb{E}[X_t]= e^{frac{1}{2} t}
$$
Hence,
$$ mathbb{E}[X_2]= e $$
answered 5 hours ago
RafaelCRafaelC
1363
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
5 hours 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, $mathbb{E}[e^X] = e^{mu + frac{1}{2} sigma^2}$ holds whenever $X sim mathcal{N}(mu, sigma^2)$
$endgroup$
– RafaelC
5 hours ago
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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oldest
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$begingroup$
Assuming you are talking about unconditional expectation, in general you have
$$
mathbb{E}[X_t] = mathbb{E}[e^{W_t}] = e^{mathbb{E}[W_t] + frac{1}{2}text{Var}(W_t) }
$$
which yields
$$
mathbb{E}[X_t]= e^{frac{1}{2} t}
$$
Hence,
$$ mathbb{E}[X_2]= e $$
answered 5 hours ago
RafaelCRafaelC
1363
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
5 hours 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, $mathbb{E}[e^X] = e^{mu + frac{1}{2} sigma^2}$ holds whenever $X sim mathcal{N}(mu, sigma^2)$
$endgroup$
– RafaelC
5 hours ago
add a comment |
$begingroup$
Assuming you are talking about unconditional expectation, in general you have
$$
mathbb{E}[X_t] = mathbb{E}[e^{W_t}] = e^{mathbb{E}[W_t] + frac{1}{2}text{Var}(W_t) }
$$
which yields
$$
mathbb{E}[X_t]= e^{frac{1}{2} t}
$$
Hence,
$$ mathbb{E}[X_2]= e $$
answered 5 hours ago
RafaelCRafaelC
1363
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
5 hours 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, $mathbb{E}[e^X] = e^{mu + frac{1}{2} sigma^2}$ holds whenever $X sim mathcal{N}(mu, sigma^2)$
$endgroup$
– RafaelC
5 hours ago
add a comment |
$begingroup$
Assuming you are talking about unconditional expectation, in general you have
$$
mathbb{E}[X_t] = mathbb{E}[e^{W_t}] = e^{mathbb{E}[W_t] + frac{1}{2}text{Var}(W_t) }
$$
which yields
$$
mathbb{E}[X_t]= e^{frac{1}{2} t}
$$
Hence,
$$ mathbb{E}[X_2]= e $$
answered 5 hours ago
RafaelCRafaelC
1363
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
$$
mathbb{E}[X_t] = mathbb{E}[e^{W_t}] = e^{mathbb{E}[W_t] + frac{1}{2}text{Var}(W_t) }
$$
which yields
$$
mathbb{E}[X_t]= e^{frac{1}{2} t}
$$
Hence,
$$ mathbb{E}[X_2]= e $$
answered 5 hours ago
RafaelCRafaelC
1363
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 5 hours ago
RafaelCRafaelC
1363
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 5 hours ago
RafaelCRafaelC
1363
answered 5 hours ago
RafaelCRafaelC
1363
1363
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
5 hours 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, $mathbb{E}[e^X] = e^{mu + frac{1}{2} sigma^2}$ holds whenever $X sim mathcal{N}(mu, sigma^2)$
$endgroup$
– RafaelC
5 hours 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
5 hours 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, $mathbb{E}[e^X] = e^{mu + frac{1}{2} sigma^2}$ holds whenever $X sim mathcal{N}(mu, sigma^2)$
$endgroup$
– RafaelC
5 hours 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
5 hours 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
5 hours 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, $mathbb{E}[e^X] = e^{mu + frac{1}{2} sigma^2}$ holds whenever $X sim mathcal{N}(mu, sigma^2)$
$endgroup$
– RafaelC
5 hours ago
$begingroup$
@Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbb{E}[e^X] = e^{mu + frac{1}{2} sigma^2}$ holds whenever $X sim mathcal{N}(mu, sigma^2)$
$endgroup$
– RafaelC
5 hours ago
add a comment |
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