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Expectation in a stochastic differential equation

3

1

$begingroup$

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

share|improve this question

edited Mar 31 at 20:09

Victor

asked Mar 31 at 19:12

VictorVictor

1016

$endgroup$

add a comment | 

3

1

$begingroup$

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

share|improve this question

edited Mar 31 at 20:09

Victor

asked Mar 31 at 19:12

VictorVictor

1016

$endgroup$

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

share|improve this question

edited Mar 31 at 20:09

Victor

asked Mar 31 at 19:12

VictorVictor

1016

$endgroup$

share|improve this question

edited Mar 31 at 20:09

Victor

asked Mar 31 at 19:12

VictorVictor

1016

share|improve this question

edited Mar 31 at 20:09

Victor

asked Mar 31 at 19:12

VictorVictor

1016

asked Mar 31 at 19:12

VictorVictor

1016

asked Mar 31 at 19:12

VictorVictor

1016

1 Answer
1

active

oldest

votes

3

$begingroup$

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,

$$ mathbbE[X_2]= e $$

share|improve this answer

answered Mar 31 at 19:52

RafaelCRafaelC

1463

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
  $endgroup$
  – RafaelC
  Mar 31 at 20:06
  

  
  
  
  

add a comment | 

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3

$begingroup$

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,

$$ mathbbE[X_2]= e $$

share|improve this answer

answered Mar 31 at 19:52

RafaelCRafaelC

1463

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
  $endgroup$
  – RafaelC
  Mar 31 at 20:06
  

  
  
  
  

add a comment | 

3

$begingroup$

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,

$$ mathbbE[X_2]= e $$

share|improve this answer

answered Mar 31 at 19:52

RafaelCRafaelC

1463

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
  $endgroup$
  – RafaelC
  Mar 31 at 20:06
  

  
  
  
  

add a comment | 

$begingroup$

Assuming you are talking about unconditional expectation, in general you have

$$
mathbbE[X_t] = mathbbE[e^W_t] = e^mathbbE[W_t] + frac12textVar(W_t)
$$

which yields

$$
mathbbE[X_t]= e^frac12 t
$$

Hence,

$$ mathbbE[X_2]= e $$

share|improve this answer

answered Mar 31 at 19:52

RafaelCRafaelC

1463

$endgroup$

share|improve this answer

answered Mar 31 at 19:52

RafaelCRafaelC

1463

answered Mar 31 at 19:52

RafaelCRafaelC

1463

answered Mar 31 at 19:52

RafaelCRafaelC

1463