Exponential of Gaussian Distribution

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SUMMARY

The expected value of an exponential Gaussian, defined as Y = exp(jX) where X follows a normal distribution N(μ, σ²), is given by E[Y] = exp(j²σ²/2 + jμ). To derive this, one can evaluate the integral E[Y] = ∫ exp(jx)f(x)dx, where f(x) is the normal density function. The Law of the Unconscious Statistician provides a systematic approach to find the expected value of a function of a random variable, specifically E[e^{jX}] = (1/σ√(2π)) ∫ e^{jx} e^{-(x-μ)²/(2σ²)} dx. Additionally, exponentials of Gaussians are associated with lognormal distributions, where if Y = exp(X), then Y has a lognormal distribution.

PREREQUISITES
  • Understanding of normal distribution N(μ, σ²)
  • Familiarity with expected value calculations
  • Knowledge of characteristic functions in probability
  • Basic integration techniques in calculus
NEXT STEPS
  • Study the Law of the Unconscious Statistician in detail
  • Learn about characteristic functions and their applications in probability
  • Explore lognormal distributions and their properties
  • Practice calculating expected values using integration techniques
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Statisticians, data scientists, mathematicians, and anyone involved in probability theory and statistical analysis will benefit from this discussion.

SeriousNoob
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I'm looking for the expected value of an exponential Gaussian

Y=\text{exp}(jX) \text{ where } X\text{~}N(\mu,\sigma^2)

From wolframalpha, http://www.wolframalpha.com/input/?i=expected+value+of+exp%28j*x%29+where+x+is+gaussian

E[Y]=\text{exp}(j^2\sigma^2/2+j\mu)

If I were to use the expected value definition:
E[Y]=\int_{-\infty}^\infty uf_Y(u)du
then I would have to figure out the pdf of Y.

I'm having trouble remembering how to get the pdf of Y, is there a more explicit way to derive the expected value?
 
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SeriousNoob said:
I'm looking for the expected value of an exponential Gaussian

Y=\text{exp}(jX) \text{ where } X\text{~}N(\mu,\sigma^2)

From wolframalpha, http://www.wolframalpha.com/input/?i=expected+value+of+exp%28j*x%29+where+x+is+gaussian

E[Y]=\text{exp}(j^2\sigma^2/2+j\mu)

If I were to use the expected value definition:
E[Y]=\int_{-\infty}^\infty uf_Y(u)du
then I would have to figure out the pdf of Y.

I'm having trouble remembering how to get the pdf of Y, is there a more explicit way to derive the expected value?
You can do it directly by evaluating ∫exp(ix)f(x)dx where f(x) is the normal density function. Also note that this is simply φ(1) where φ(t) is the characteristic function of this particular normal distribution.
 
SeriousNoob said:
I'm looking for the expected value of an exponential Gaussian

Y=\text{exp}(jX) \text{ where } X\text{~}N(\mu,\sigma^2)

From wolframalpha, http://www.wolframalpha.com/input/?i=expected+value+of+exp%28j*x%29+where+x+is+gaussian

E[Y]=\text{exp}(j^2\sigma^2/2+j\mu)

If I were to use the expected value definition:
E[Y]=\int_{-\infty}^\infty uf_Y(u)du
then I would have to figure out the pdf of Y.

I'm having trouble remembering how to get the pdf of Y, is there a more explicit way to derive the expected value?

What mathman said is right - I just wanted to add a couple of things:

1) The Law of the Unconscious Statistician is a snarky name for the usual way to find the EV of a function of a random variable.

##
E[y(X)] = \int y(x) f_X(x) dx
##
##
E[e^{jX}] = \frac{1}{\sigma\sqrt{2\pi}} \int e^{jx} e^{-\frac{(x-\mu)^2}{2 \sigma^2}} dx
##

2) Exponentials of Gaussians show up often enough to have their own name. If ##Y = \exp(X)##, then ##Y## has a lognormal distribution. The name 'lognormal' is a reminder that ##\log(Y)## is normally distributed.
 
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