Probability- Expected value of e^x

[Roni1985]

Homework Statement

Find E[e^x] where x~N($$\mu$$, sigma squared)

Homework Equations

The Attempt at a Solution

It looks like a moment generating function.
Here is what I did:
Assume X= $$\mu$$ + $$\sigma$$*Z

E[e^tx]= E[e^{t($$\mu$$+$$\sigma$$*Z)}]

I simplified it and used the fact of moment generating functions and got
=exp{$$\sigma$$²*t²/2+$$\mu$$*t}
I plugged in t=1 and that was my answer.

Do you think it makes sense?
Is there a better/faster way?

Thanks.

 

[Stephen Tashi]

Your statement of the problem appear garbled. Did it just say "Find the expectation of e^x", without telling you how x was distributed?

You say that you assume x is normally distributed. Why did you write it as
$$\mu + \sigma * N(\mu,\sigma_2)$$ ?

It's just a general to write it as $$\mu + \sigma N(0,1)$$

 

[Roni1985]

Your statement of the problem appear garbled. Did it just say "Find the expectation of e^x", without telling you how x was distributed?

You say that you assume x is normally distributed. Why did you write it as
$$\mu + \sigma * N(\mu,\sigma_2)$$ ?

It's just a general to write it as $$\mu + \sigma N(0,1)$$

You are totally right... hold on... editing my question

 

[Stephen Tashi]

Here is what I did:
Assume X= $$\mu$$ + $$\sigma$$*Z
E[e^tx]= E[e^{t($$\mu$$+$$\sigma$$*Z)}]

I simplified it and used the fact of moment generating functions

Explain what you mean by "the fact of moment generating functions".

and got
=exp{$\sigma$²*t²/2+$\mu$*t}


I plugged in t=1 and that was my answer.

Do you think it makes sense?

It doesn't make sense to me. You didn't really say what your answer was.

 

[Roni1985]

Explain what you mean by "the fact of moment generating functions".
It doesn't make sense to me. You didn't really say what your answer was.

This is my final answer:
exp{sigma^2/2+$$\mu$$}

How would you approach this question?

 

[Stephen Tashi]

The moment generating function for the random variable

$$e^x = e^{(\mu + \sigma z)}$$

is

$$E( e^{t e^{\mu + \sigma z}})$$

which looks difficult to compute.

It seems simpler to compute the expected value from the definition of expected value.

$$E( e^x) = E(e^{\mu + \sigma z}) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{\mu + \sigma z} e^{ \frac{-z^2}{2} } dz$$

 

[Roni1985]

The moment generating function for the random variable

$$e^x = e^{(\mu + \sigma z)}$$

is

$$E( e^{t e^{\mu + \sigma z}})$$

which looks difficult to compute.

It seems simpler to compute the expected value from the definition of expected value.

$$E( e^x) = E(e^{\mu + \sigma z}) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{\mu + \sigma z} e^{ \frac{-z^2}{2} } dz$$

Got it... the moment generating function of a normally distributed r.v. solves the same integral.

Thanks very much for the help... appreciate it.

Roni.

 

[Stephen Tashi]

Got it... the moment generating function of a normally distributed r.v. solves the same integral.

You'll have to explain that to me Roni, I don't see it.

And, by the way, I don't think you substitute t = 1 into the moment generating function of a random variable to get the mean of the random variable.