Question

Log-normal probability distribution A commonly used distribution in probability and statistics is the log-normal distribution. (If the logarithm of a variable has a normal distribution, then the variable itself has a log-normal distribution.) The distribution function is
f(x) = 1/xσ√(2π) e⁻ˡⁿ^² ˣ / ²σ^², for x ≥ 0
where ln x has zero mean and standard deviation σ > 0.
b. Evaluate lim x → 0 ƒ(x). (Hint: Let x = eʸ.)

Accepted Answer

Start by writing down the given probability density function (pdf) for the log-normal distribution:  
$$f(x) = \frac{1}{x \sigma \sqrt{2\pi}} e$$^{-\frac{(\ln x)^2}$${2 \$$sigma^2$$}}$$, for $$x \geq 0. To $$evaluate the limit as $$x$$ approaches 0, use the hint and substitute $$x = e^y. $$This means as $$x 	o 0^+$$, we have $$y = \ln x 	o -\infty. $$Rewrite the function in terms of $$y$$:  
$$f(e^y) = \frac{1}{e^y \sigma \sqrt{2\pi}} e$$^{-\frac{y^2}$${2 \$$sigma^2$$}} = \frac{1}{\sigma \sqrt{2\pi}} e$$^{-y}$$ e$$^{-\frac{y^2}$${2 \$$sigma^2$$}}. $$Combine the exponents to get a single exponential expression:  
$$f(e^y) = \frac{1}{\sigma \sqrt{2\pi}} e$$^{-y - \frac{y^2}$${2 \$$sigma^2$$}}. $$Analyze the behavior of the exponent $$-y - \frac{y^2$}{2 \$$sigma^2$$} as$$ $$y 	o -\infty. $$This will help determine the limit of $$f(x) as$$ $$x 	o 0^+.$$

Verified video answer for a similar problem:

Key Concepts

Log-normal Distribution

Limit Evaluation Using Substitution

Behavior of Exponential and Logarithmic Functions at Infinity