A random variable has probability density function , . (1) Determine the value of . (2) Find the distribution of . (3) Discuss whether is a continuous random vector.

1. . 2. , with density . 3. The values of are almost surely confined to the curve , which violates the definition of a continuous random vector; therefore is not a continuous random vector.

Probability Theory Solution – Problem 8: Determine the value of

Question

A random variable $X$ has probability density function $p(x)=\frac{c}{x}e^{-\frac{(\ln x)^{2}}{2}}$ , $x>0$ . (1) Determine the value of $c$ . (2) Find the distribution of $\ln X$ . (3) Discuss whether $(X,\ \ln X)$ is a continuous random vector.

Step-by-step solution

Step 1. Set up the normalisation integral. The given density is $p(x)=\dfrac{c}{x}e^{-\dfrac{(\ln x)^2}{2}}$ , $x>0$ . A necessary condition for this to be a probability density function is $\int_{0}^{+\infty}p(x)\,dx=1.$

Substituting $p(x)$ yields $\int_{0}^{+\infty}\dfrac{c}{x}e^{-\dfrac{(\ln x)^2}{2}}\,dx=1.$

Step 2. Evaluate the integral by substitution.

Let $t=\ln x$ , so that $x=e^t$ and $dx=e^t dt$ . Then $\dfrac{dx}{x}=\dfrac{e^t dt}{e^t}=dt.$

As $x$ ranges from $0$ to $+\infty$ , $t=\ln x$ ranges from $-\infty$ to $+\infty$ . Hence

$\int_{0}^{+\infty}\dfrac{c}{x}e^{-\dfrac{(\ln x)^2}{2}}\,dx =c\int_{0}^{+\infty}\dfrac{1}{x}e^{-\dfrac{(\ln x)^2}{2}}\,dx =c\int_{-\infty}^{+\infty}e^{-\dfrac{t^2}{2}}\,dt.$

It is well known that $\int_{-\infty}^{+\infty}e^{-\dfrac{t^2}{2}}\,dt=\sqrt{2\pi}.$

Therefore the normalisation condition becomes $c\sqrt{2\pi}=1.$

Step 3. Solve for the constant $c$ .

$c=\dfrac{1}{\sqrt{2\pi}}.$

Hence the value of $c$ is $\dfrac{1}{\sqrt{2\pi}}$ .

Step 1. Obtain the density via the change-of-variable formula.

Let $Y=\ln X$ and write $g(x)=\ln x$ , whose inverse is $x=e^y$ . Since $X>0$ , the range of $Y$ is the entire real line, i.e. $y\in(-\infty,+\infty)$ .

For a continuous random variable, the one-dimensional invertible transformation formula states: if $Y=g(X)$ and $g$ is monotone and invertible, then $f_Y(y)=f_X(x)\left|\dfrac{dx}{dy}\right|\bigg|_{x=g^{-1}(y)}.$

Here $x=e^y$ and $\dfrac{dx}{dy}=e^y$ , so $f_Y(y)=f_X(e^y)\cdot e^y.$

Step 2. Substitute the known density $f_X$ and simplify.

From Part (1), $f_X(x)=\dfrac{1}{\sqrt{2\pi}}\dfrac{1}{x}e^{-\dfrac{(\ln x)^2}{2}},\quad x>0.$

Setting $x=e^y$ gives

$f_Y(y)=\dfrac{1}{\sqrt{2\pi}}\dfrac{1}{e^y}e^{-\dfrac{(\ln e^y)^2}{2}}\times e^y =\dfrac{1}{\sqrt{2\pi}}e^{-\dfrac{y^2}{2}},\quad y\in\mathbb{R}.$

This is precisely the probability density function of the standard normal distribution.

Step 3. State the conclusion.

Therefore $Y=\ln X$ follows the standard normal distribution, i.e. $\ln X\sim \text{N}(0,1).$

Equivalently, the distribution function is $P(\ln X\le y)=\Phi(y),\quad y\in\mathbb{R},$ where $\Phi(y)$ denotes the standard normal distribution function.

Step 1. Recall the definition of a continuous two-dimensional random vector.

A random vector $(U,V)$ is called a continuous random vector if there exists a non-negative integrable function $f_{U,V}(u,v)$ such that for every measurable set $B\subset\mathbb{R}^2$ , $P\big((U,V)\in B\big)=\iint_B f_{U,V}(u,v)\,du\,dv,$ and $\iint_{\mathbb{R}^2} f_{U,V}(u,v)\,du\,dv=1.$

Under this definition, if $f_{U,V}$ exists, then for any set $B$ of planar Lebesgue measure zero, $P\big((U,V)\in B\big)=0.$

Step 2. Analyse the support of $(X,\ln X)$ .

Set $Y=\ln X$ . Then $P\big(Y=\ln X\big)=1,$ meaning the random vector $(X,Y)$ satisfies $y=\ln x$ almost surely.

Consequently, the values of $(X,Y)$ are almost surely contained in the set $A=\big\{(x,y)\in\mathbb{R}^2:\ x>0,\ y=\ln x\big\},$ i.e. $P\big((X,Y)\in A\big)=1.$

However, $A$ is a smooth curve in the plane; as a one-dimensional curve in $\mathbb{R}^2$ , its two-dimensional Lebesgue measure is zero.

Step 3. Compare with the definition to reach the conclusion.

If $(X,Y)$ were a continuous random vector, then for any set $B$ of Lebesgue measure zero we would have $P\big((X,Y)\in B\big)=0$ . Yet here there exists a set $A$ of measure zero with $P\big((X,Y)\in A\big)=1$ .

This contradicts the properties of a continuous random vector. Therefore: $(X,\ \ln X)$ is not a continuous random vector.

Final answer

1. $c=\dfrac{1}{\sqrt{2\pi}}$ . 2. $\ln X\sim \text{N}(0,1)$ , with density $f_{\ln X}(y)=\dfrac{1}{\sqrt{2\pi}}e^{-\dfrac{y^2}{2}}$ . 3. The values of $(X,\ \ln X)$ are almost surely confined to the curve $y=\ln x$ , which violates the definition of a continuous random vector; therefore $(X,\ \ln X)$ is not a continuous random vector.

Marking scheme

The following is the complete marking scheme for this probability theory problem (full marks: 7 points).

1. Checkpoints (Total 7 pts)

Part 1: Determining the value of $c$ (2 points)

*Note: This part tests the ability to find a constant using the normalisation property of integrals.*

Setting up the integral and performing the substitution [1 pt]
State the normalisation condition $\int_{0}^{+\infty} \frac{c}{x}e^{-\frac{(\ln x)^2}{2}} dx = 1$ and carry out the substitution $t=\ln x$ (or $dx/x = dt$ ), converting the integral into the Gaussian integral form $\int_{-\infty}^{+\infty} e^{-t^2/2} dt$ .
*Note: If the substitution is not shown explicitly but the student directly invokes the definition of the log-normal distribution and correctly identifies the normalisation constant, full credit may still be awarded.*
Computing the result [1 pt]
Correctly obtain $c = \frac{1}{\sqrt{2\pi}}$ .

Part 2: Finding the distribution of $\ln X$ (3 points)

*Note: Grade one of the following paths only | scores across paths are not cumulative. This part tests distribution transformation of a function of a random variable.*

Path A: Density transformation method
Jacobian/derivative term [1 pt]: Write down the inverse $x=e^y$ of the transformation $y=\ln x$ together with its derivative (or Jacobian determinant) $\frac{dx}{dy} = e^y$ .
Substitution and simplification [1 pt]: Correctly substitute $x=e^y$ and the derivative into the density transformation formula $f_Y(y) = p(e^y) \cdot e^y$ and simplify to obtain $c e^{-y^2/2}$ or $\frac{1}{\sqrt{2\pi}} e^{-y^2/2}$ .
Final conclusion [1 pt]: Explicitly state that $\ln X$ follows the standard normal distribution $N(0,1)$ , or write out the complete standard normal probability density function (including the domain $y\in\mathbb{R}$ ).
Path B: Distribution function method
Definition and conversion [1 pt]: Write the distribution function $F_Y(y) = P(\ln X \le y) = P(X \le e^y) = \int_{0}^{e^y} p(x) dx$ .
Integral substitution [1 pt]: Use the substitution $t=\ln x$ to convert the limits and integrand into the standard normal distribution function form $\int_{-\infty}^{y} c e^{-t^2/2} dt$ .
Final conclusion [1 pt]: Identify the integral as the standard normal distribution; conclusion same as Path A.
Shared prerequisite [max 1 pt]: If the student made an error in Part 1 leading to an incorrect value of $c$ , but the reasoning in this part is entirely correct, only the result mark is deducted; process marks are retained (follow-through marking).

Part 3: Discussing whether $(X, \ln X)$ is a continuous random vector (2 points)

*Note: This part tests understanding of the definition of a two-dimensional continuous random vector.*

Identifying the support set / measure [1 pt]
Point out that the probability mass of the random vector $(X, \ln X)$ is concentrated on the plane curve $y = \ln x$ ; or note that its support has two-dimensional Lebesgue measure zero.
Stating the conclusion [1 pt]
Based on the above reasoning (a set of measure zero carries probability one, or no joint density $f(x,y)$ with respect to two-dimensional Lebesgue measure can exist), conclude that $(X, \ln X)$ is not a continuous random vector.

Total (max 7)

2. Zero-credit items

Copying the problem statement: Merely restating the given conditions or formula names (e.g. "use the density formula") without performing any concrete substitution or computation.
Unsupported guess in Part 3: Answering only "yes" or "no" in Part 3 without any mathematical justification.
Conceptual confusion: Arguing in Part 3 that "because the marginal distributions of $X$ and $\ln X$ are both continuous, the joint distribution is also continuous" (this conclusion is false; award 0 points).
Incorrect independence assumption: Attempting in Part 3 to construct a density via $f(x,y) = f_X(x) \cdot f_Y(y)$ (independence is not given, and the variables are in fact perfectly dependent; this approach receives 0 points).

3. Deductions

*Apply at most one of the following deductions (whichever is most severe); the total score cannot fall below 0.*

Computational/arithmetic error (-1 pt):
Errors in handling constants during integration (e.g. omitting $\sqrt{2\pi}$ , miscalculating coefficients), leading to an incorrect value of $c$ or incorrect distribution parameters.
Missing domain (-1 pt):
When writing the probability density function as a final answer, failing to specify the range of the variable (e.g. $y \in \mathbb{R}$ or $-\infty < y < +\infty$ ). This deduction is waived if the student explicitly writes "follows the standard normal distribution" in words.
Logical/notational confusion (-1 pt):
Severely confusing the random variable notation (uppercase $X$ ) with the realisation notation (lowercase $x$ ), or writing logically incoherent integration limits (e.g. $0$ to $\ln x$ ) rendering the mathematical expression meaningless.

Probability Theory – Problem 8: Determine the value of