After completing the square, the integral becomes the integral of a normal PDF with mean + ² and variance ². Since the integral of any PDF over its entire domain is 1, the expected value of Y simplifies to the exponential term outside the integral.

After completing the square, the integral becomes the integral of a normal PDF with mean + ² and variance ². Since the integral of any PDF over its entire domain is 1, the expected value of Y simplifies to the exponential term outside the integral.

Using the general result that if X N(, ²), then E[e^(aX)] = e^(a + (a² ²)/(2)), we substitute a=2. This gives us the value for E[Y²].

The variance of Y is defined as E[Y²] - (E[Y])². We substitute the expressions we found for E[Y²] and E[Y] and simplify the result by factoring out common terms.

Let X ~ N(μ, σ²) and $Y = e^X \text{(log-\text{normal distribution})}$ . Find E[Y] and Var(Y).

Answer: $E[Y] = e^{\mu + \frac{\sigma^2}{2}}$ $\text{Var}(Y) = e^{2\mu + \sigma^2}(e^{\sigma^2} - 1)$

Step-by-step solution

Step 1: Define the probability density function of X

Since $X$ follows a normal distribution with mean $\mu$ and variance $\sigma^2$ , its probability density function (PDF) is given by this formula. This PDF will be used to calculate the expected value of $Y$ .

Step 2: Calculate E[Y] using the definition of expectation

The expected value of a function of a random variable, $E[g(X)]$ , is found by integrating $g(x)$ multiplied by the PDF of $X$ over its entire range. Here, $g(X) = e^X$ , so we substitute $e^x$ and the PDF of $X$ into the integral.

Step 3: Simplify the exponent and complete the square

We combine the exponential terms and complete the square in the exponent. This involves algebraic manipulation to transform the expression into the form of a normal distribution's PDF. The integral of a PDF over its entire domain is 1, which simplifies the expression.

Step 4: Calculate E[Y]

After completing the square, the integral becomes the integral of a normal PDF with mean $\mu + \sigma^2$ and variance $\sigma^2$ . Since the integral of any PDF over its entire domain is 1, the expected value of $Y$ simplifies to the exponential term outside the integral.

Step 5: Calculate E[Y²]

To find the variance, we first need to calculate $E[Y^2]$ . Since $Y = e^X$ , $Y^2 = (e^X)^2 = e^{2X}$ . We can use the same method as for $E[Y]$ , but with $2X$ instead of $X$ . This is equivalent to a normal distribution with mean $2\mu$ and variance $4\sigma^2$ .

Step 6: Apply the formula for E[e^(aX)]

Using the general result that if $X \sim N(\mu, \sigma^2)$ , then $E[e^{aX}] = e^{a\mu + \frac{a^2\sigma^2}{2}}$ , we substitute $a=2$ . This gives us the value for $E[Y^2]$ .

Step 7: Calculate Var(Y)

The variance of $Y$ is defined as $E[Y^2] - (E[Y])^2$ . We substitute the expressions we found for $E[Y^2]$ and $E[Y]$ and simplify the result by factoring out common terms.

Solve your own maths question free →

Let X ~ N(μ, σ²) and Y=eX(log-normal distribution)Y = e^X \text{(log-\text{normal distribution})}Y=eX(log-normal distribution). Find E[Y] and Var(Y).