Expected Value of Exponential of Normal

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Problem Statement

If X is a normal distribution X \sim N(\mu, \sigma^2), what is E[e^X]?

Solution

This problem requires computing the expected value of an exponential function of a normal random variable. This is a fundamental result that connects to the moment-generating function and has important applications in finance, particularly in modeling log-normal distributions.

Step 1: Set Up the Expectation

For a continuous random variable X with PDF f(x), the expected value of g(X) is:

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

In our case, g(X) = e^X and X \sim N(\mu, \sigma^2), so:

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

Step 2: Use the Moment-Generating Function (MGF)

The most elegant approach uses the moment-generating function of the normal distribution. For X \sim N(\mu, \sigma^2), the MGF is:

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

Setting t = 1:

E[e^X] = M_X(1) = \exp\left(\mu \cdot 1 + \frac{\sigma^2 \cdot 1^2}{2}\right) = \exp\left(\mu + \frac{\sigma^2}{2}\right)

Step 3: Alternative Derivation Using Completing the Square

We can also derive this directly by completing the square in the exponent. Starting with:

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

Combining the exponentials:

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

Now we complete the square in the exponent. Let’s work with:

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

Completing the square:

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

After simplification, this becomes:

= -\frac{(x - (\mu + \sigma^2))^2}{2\sigma^2} + \mu + \frac{\sigma^2}{2}

Therefore:

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

The integral is the PDF of N(\mu + \sigma^2, \sigma^2) integrated over its entire support, which equals 1. Therefore:

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

Final Answer

For X \sim N(\mu, \sigma^2):

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

Special Cases

Case 1: Standard Normal (\mu = 0, \sigma^2 = 1)

E[e^X] = e^{0 + \frac{1}{2}} = e^{1/2} = \sqrt{e} \approx 1.649

Case 2: Mean Zero (\mu = 0)

E[e^X] = e^{\frac{\sigma^2}{2}}

This shows that even when the mean is zero, the expected exponential is greater than 1, and it increases with variance.

Key Insights

  1. Jensen’s Inequality: Since e^x is a convex function, by Jensen’s inequality, E[e^X] \geq e^{E[X]} = e^{\mu}. Our result e^{\mu + \sigma^2/2} \geq e^{\mu} confirms this, with the extra \sigma^2/2 term accounting for the variance.

  2. Moment-Generating Function: This result is a direct application of the MGF of the normal distribution, which is one of the most elegant properties of normal distributions.

  3. Log-Normal Distribution: If X \sim N(\mu, \sigma^2), then Y = e^X follows a log-normal distribution with parameters \mu and \sigma^2. The mean of the log-normal is E[Y] = e^{\mu + \sigma^2/2}, which is exactly our result.

  4. Variance Effect: The variance term \sigma^2/2 in the exponent shows that higher variance increases E[e^X] exponentially. This is important in finance where volatility affects expected returns.

  5. Symmetry Breaking: Even if \mu = 0 (symmetric distribution), E[e^X] > 1 because the exponential function is convex and amplifies positive values more than it reduces negative values.

Applications

  1. Finance - Black-Scholes Model: In option pricing, stock prices are often modeled as S_t = S_0 e^{X} where X is normally distributed. The expected stock price uses this formula.

  2. Log-Normal Models: When modeling quantities that must be positive (prices, populations, etc.), the log-normal distribution (exponential of normal) is commonly used.

  3. Risk Management: Understanding how volatility (\sigma^2) affects expected exponential values is crucial for risk assessment.

  4. Monte Carlo Simulation: When simulating log-normal processes, this formula helps verify simulation accuracy.

  5. Geometric Brownian Motion: In stochastic processes, this result appears when computing expected values of geometric Brownian motion.

Relationship to Other Concepts

  1. Moment-Generating Function: This is M_X(1) where M_X(t) = E[e^{tX}] is the MGF.

  2. Characteristic Function: Related to the characteristic function \phi_X(t) = E[e^{itX}] for complex t.

  3. Log-Normal Distribution: If Y = e^X where X \sim N(\mu, \sigma^2), then Y is log-normal with E[Y] = e^{\mu + \sigma^2/2}.

  4. Jensen’s Inequality: Demonstrates that E[e^X] > e^{E[X]} for non-degenerate normal distributions.

Common Pitfalls

  1. Forgetting the variance term: A common mistake is to think E[e^X] = e^{\mu}, missing the crucial \sigma^2/2 term.

  2. Confusing with e^{E[X]}: E[e^X] \neq e^{E[X]} due to Jensen’s inequality. The correct formula is E[e^X] = e^{\mu + \sigma^2/2} > e^{\mu}.

  3. Sign errors: Be careful with the sign in the exponent. It’s \mu + \sigma^2/2, not \mu - \sigma^2/2.

  4. Variance vs Standard Deviation: Remember that \sigma^2 is the variance, not the standard deviation. If given \sigma (standard deviation), use \sigma^2 in the formula.

Summary

For X \sim N(\mu, \sigma^2):

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

This fundamental result connects the normal distribution to the log-normal distribution and has wide applications in finance, risk management, and stochastic modeling. The variance term \sigma^2/2 in the exponent is crucial and reflects the convexity of the exponential function.