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Thread: Expected value of e^(normal random variable)

  1. #1
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    Expected value of e^(normal random variable)

    Hi everyone,

    modelling stock prices I have the following generalized Wiener process:

    dS_t=S_t(\mu d_t + \sigma dW_t)

    Applying It's lemma gives us:

    ln(S_t)-ln(S_0)=(\mu-\frac{\sigma^2}{2})t+\sigma\epsilon\sqrt{t}

    with \epsilon being a standard normal random variable with mean 0 and variance 1. Taking exp of the above, gives us:

    S_t=S_0 exp\left(\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma\epsilon\sqrt{t} \right)

    That gives us the expected value of S_t which I don't understand.

    E[S_t]=S_0exp(\mu t})

    Specifically my question is: Why is the expected value of

    E[e^X]=e^{\mu +\frac{1}{2}\sigma^2} \qquad ?

    with X being a normal random variable with mean \mu and variance \sigma^2.

    Thank you for explanation/proof, but also for relevant literature.
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  2. #2
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    Re: Expected value of e^(normal random variable)

    Hey neptunhiker.

    Have you tried deriving the PDF for a log-normal distribution?

    Hint: You can do it with the transformation theorem where if Y = h(X) and h is invertible (i.e. has an inverse) you get get the PDF of Y given that you have the PDF of X.

    Once you do this, you can derive the results for the mean and the variance.

    Alternatively you can use a Normal PDF and use the formula E[e^X] = Integral (over Real Line) e^x * f(x) dx (and similarly for the variance).
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  3. #3
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    Re: Expected value of e^(normal random variable)

    Hi chiro,

    Thank you for your reply. Unfortunately I don't follow.

    If X\sim \phi(0,1) then its density function is given by:

    f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}

    and therefore:

    E[X]=\int_{-\infty}^{\infty}x\cdot f(x)\,dx \qquad \text{right?}

    Then is also true:

    E[e^X]=\int_{-\infty}^{\infty}e^x\cdot f(x)\,dx \qquad \text{right?}

    So far, I think I have followed your second suggestion. But what now? I am stuck here. I hope you can help me out. Thanks.
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  4. #4
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    Re: Expected value of e^(normal random variable)

    You have to complete the square for the exponential term and then use the fact that Integral 1/SQRT(2*pi) * e^-(x-mu)^2/2sigma^2 = 1 for the real line for any mu and non-zero sigma.

    Factor out the other term and you should get your answer.

    Hint: Recall that (x+c)^2 = x^2 + 2cx + c^2.
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  5. #5
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    Re: Expected value of e^(normal random variable)

    Hi Chiro,

    Thanks again. I just don't know how to complete the square for the exponential term. My knowledge seems to just be limited in that point. I understand the following:

    x^2-6x+17=x^2-6x+9-9+17=(x-3)^2-9+17=(x-3)^2+8

    I guess I am on the right track, but working with integrals just sets my limits. I am still hoping for mercy and a little more help

    Thanks.
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  6. #6
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    Re: Expected value of e^(normal random variable)

    Hint: -(1/2x^2 - x + 2 - 2) = -(x/SQRT(2) - SQRT(2))^2 + 2 = [-(x - 2)^2]/2 + 2.

    This implies that your integral = e^2 * Integral [-infinity,infinity] * 1/SQRT(2*pi*1) * e^([-(x-2)]^2]/2 = e^2.
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  7. #7
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    Re: Expected value of e^(normal random variable)

    Hello Chiro,

    following your hint:

    Hint: -(1/2x^2 - x + 2 - 2) = -(x/SQRT(2) - SQRT(2))^2 + 2 = [-(x - 2)^2]/2 + 2.
    So, you are saying:

    -\left(\frac{1}{2}x^2-x+2-2 \right)=-\left(\frac{x}{\sqrt{2}}-\sqrt{2} \right)^2+2=\frac{-(x-2)^2}{2}+2

    First of all, I believe that you meant

    -\left(\frac{1}{2}x^2\textbf{-2x}+2-2 \right)=-\left(\frac{x}{\sqrt{2}}-\sqrt{2} \right)^2+2=\frac{-(x-2)^2}{2}+2 \qquad \text{right?}

    Then you are saying:

    This implies that your integral = e^2 * Integral [-infinity,infinity] * 1/SQRT(2*pi*1) * e^([-(x-2)]^2]/2 = e^2.
    therefore:

    e^2\cdot \int^{\infty}_{-\infty}\frac{1}{\sqrt{2\pi\cdot1}}\cdot exp\left[\frac{(-(x-2))^2}{2}\right]=e^2

    Sorry, but I don't follow at all. How is that helping me solving the following:

    E[e^X]=\int^{\infty}_{-\infty}e^x\cdot f(x)\,dx

    with

    f(x)=\frac{1}{\sqrt{2\pi \sigma^2}}exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right]
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  8. #8
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    Re: Expected value of e^(normal random variable)

    What I mean is that you have to transform your integral into one that looks like a normal PDF and then use the fact the integral of that PDF is 1.

    You will after you complete the square and take the terms out (in the way that I showed you) you will get some constant multiplied by the integral and that will be your expected value.
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  9. #9
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    Re: Expected value of e^(normal random variable)

    Hi Chiro,

    after quite a long weekend of research and sweat I believe I have found a solution. Here we go:

    E[e^X]=\int_{-\infty}^{\infty} exp(x) \cdot f(x)\,dx

    E[e^X]=\int_{-\infty}^{\infty} exp(x) \cdot \frac{1}{\sqrt{2\pi\sigma^2}}\,exp\left(\frac{-(x-\mu)^2}{2\sigma^2}\right)\,dx

    E[e^X]=\frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{\infty}exp\left(x-\frac{(x-\mu)^2}{2\sigma^2}\right)\,dx

    E[e^X]=\frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{\infty}exp\left[-\frac{1}{2\sigma^2}\left(x^2+(-2\sigma^2-2\mu)x+\mu^2\right)\right]\,dx

    Now completing the square of the exponential term:

    E[e^X]=\frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{\infty}exp\left[-\frac{1}{2\sigma^2}\left(x^2+(-2\sigma^2-2\mu)x+(\sigma^4+2\sigma^2+\mu^2)-(\sigma^4+2\sigma^2+\mu^2)+\mu^2\right)\right]\,dx

    E[e^X]=\frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{\infty}exp\left[-\frac{1}{2\sigma^2}\left((x-(\sigma^2+\mu))^2-\sigma^4-2\sigma^2\mu\right\right]\,dx

    E[e^X]=\frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{\infty}exp\left[-\frac{1}{2\sigma^2}(x-(\sigma^2+\mu))^2+\frac{\sigma^2}{2}+\mu\right]\,dx

    E[e^X]=exp\left(\mu+\frac{\sigma^2}{2}\right)\cdot \frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{\infty}exp\left[-\frac{1}{2\sigma^2}(x-(\sigma^2+\mu))^2\right]\,dx

    And that gives us:

    E[e^X]=exp\left(\mu+\frac{\sigma^2}{2}\right)

    which implies that the following must be true:

    \frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{\infty}exp\left[-\frac{1}{2\sigma^2}(x-(\sigma^2+\mu))^2\right]\,dx=1

    We can further rewrite the above:

    \frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{\infty}exp\left[\frac{-(x+\mu)^2}{2\sigma^2}+x-\frac{\sigma^2}{2}-\mu\right]\,dx=1

    And that's where I don't get any further, because I don't understand why the above is equal to one. Basically I have two questions:

    a) Is the way I have tried to solve for E[X] correct?
    b) Is the last equation equal to one and if so, why?

    Thank you for help and answers.
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  10. #10
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    Re: Expected value of e^(normal random variable)

    The first way you figured it out is spot on and is the best way to do it.

    The second one is also correct (i.e. the first expansion not the second) if you assume that the variance is sigma^2 and that the mean = sigma^2 + mu.

    You can relabel them as mu_proper = mu + sigma^2 and sigma_proper_squared =sigma^2.

    Since they are independent quantities you just get a normal distribution with those parameters (at least you should).
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  11. #11
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    Re: Expected value of e^(normal random variable)

    Hi chiro,

    thanks a lot. I think I get it now. Here are the last steps of my final solution:

    E[e^X]=exp\left(\mu+\frac{\sigma^2}{2}\right)\cdot \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}exp\left[-\frac{1}{2\sigma^2}(x-(\sigma^2+\mu))^2\right]dx

    and then by substituting m=\sigma^2+\mu and s=\sigma^2 we will get another normal probability density function with mean m and variance s^2 as integrand and therefore end up with:

    E[e^X]=exp\left(\mu+\frac{\sigma^2}{2}\right)\cdot \underbrace{\frac{1}{\sqrt{2\pi s^2}}\int_{-\infty}^{\infty}exp\left[\frac{-(x-m)^2}{2s^2}\right]dx}_{1}

    and then after only two weeks I get the final result

    E[e^X]=exp\left(\mu+\frac{\sigma^2}{2}\right)

    chiro, without you, I wouldn't have done it. Thanks a lot
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