Expectation in a stochastic differential equation
$begingroup$
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 Mar 31 at 19:12
VictorVictor
916
$endgroup$
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 Mar 31 at 19:12
VictorVictor
916
$endgroup$
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 Mar 31 at 19:12
VictorVictor
916
$endgroup$
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 Mar 31 at 19:12
VictorVictor
916
asked Mar 31 at 19:12
VictorVictor
916
edited Mar 31 at 20:09
Victor
asked Mar 31 at 19:12
VictorVictor
916
asked Mar 31 at 19:12
VictorVictor
916
asked Mar 31 at 19:12
VictorVictor
916
916
add a comment |
add a comment |
1 Answer
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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 Mar 31 at 19:52
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
Mar 31 at 20:00
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
Mar 31 at 20:06
add a comment |
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1 Answer
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1 Answer
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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 Mar 31 at 19:52
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
Mar 31 at 20:00
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
Mar 31 at 20:06
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 Mar 31 at 19:52
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
Mar 31 at 20:00
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
Mar 31 at 20:06
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 Mar 31 at 19:52
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
$$
mathbb{E}[X_t] = mathbb{E}[e^{W_t}] = e^{mathbb{E}[W_t] + frac{1}{2}text{Var}(W_t) }
$$
which yields
$$
mathbb{E}[X_t]= e^{frac{1}{2} t}
$$
Hence,
$$ mathbb{E}[X_2]= e $$
answered Mar 31 at 19:52
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 Mar 31 at 19:52
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 Mar 31 at 19:52
RafaelCRafaelC
1463
answered Mar 31 at 19:52
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
Mar 31 at 20:00
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
Mar 31 at 20:06
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
Mar 31 at 20:00
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
Mar 31 at 20:06
$begingroup$
I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
$endgroup$
– Victor
Mar 31 at 20:00
$begingroup$
I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
$endgroup$
– Victor
Mar 31 at 20:00
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
Mar 31 at 20:06
$begingroup$
@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
Mar 31 at 20:06
add a comment |
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