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

1. ; 2. , with density ; 3. The values of are almost surely concentrated on the curve , which does not satisfy the definition of a continuous random vector; therefore it is not a continuous random vector.

Probability Theory Solution – Problem 64: Compute the value of ;

Question

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

Step-by-step solution

1. Write the normalization condition and set up the integral Given the density $p(x)=\dfrac{c}{x}e^{-\dfrac{(\ln x)^2}{2}}$ , $x>0$ , the necessary condition for it to be a probability density function is $\int_{0}^{+\infty}p(x)\,dx=1.$

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

2. Evaluate the integral via substitution

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

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

Using the known result $\int_{-\infty}^{+\infty}e^{-\dfrac{t^2}{2}}\,dt=\sqrt{2\pi},$

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

3. Solve for the constant $c$

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

1. Use the one-dimensional transformation formula to find the density

Let $Y=\ln X$ , and set $g(x)=\ln x$ , so $x=e^y$ is the inverse function. Since $X>0$ , $Y$ takes values on the entire real line, i.e., $y\in(-\infty,+\infty)$ .

The one-dimensional invertible transformation formula for continuous random variables 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 $f_Y(y)=f_X(e^y)\cdot e^y.$

2. Substitute the known $f_X$ and simplify

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

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

This is precisely the standard normal probability density function.

3. State the conclusion

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

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

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 any 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 $B$ with two-dimensional Lebesgue measure $0$ , $P\big((U,V)\in B\big)=0.$

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

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

Therefore $(X,Y)$ takes values almost entirely on 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, the set $A$ is a smooth curve in the plane; it is a one-dimensional curve in $\mathbb{R}^2$ and has two-dimensional Lebesgue measure $0$ .

3. Compare with the definition to reach the conclusion

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

This contradicts the property of continuous random vectors. 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 concentrated on the curve $y=\ln x$ , which does not satisfy the definition of a continuous random vector; therefore it is not a continuous random vector.

Marking scheme

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

1. Checkpoints (Total 7 pts)

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

*Note: This part tests the use of integral normalization to find the constant.*

Setting up the integral and substitution [1 pt]
Write the normalization condition $\int_{0}^{+\infty} \frac{c}{x}e^{-\frac{(\ln x)^2}{2}} dx = 1$ and perform the substitution $t=\ln x$ (or $dx/x = dt$ ), transforming the integral into the Gaussian integral form $\int_{-\infty}^{+\infty} e^{-t^2/2} dt$ .
*Note: If the substitution process is not explicitly shown but the normalization coefficient relationship is correctly identified using the definition of the lognormal distribution, this point may still be awarded.*
Solving for the result [1 pt]
Correctly compute $c = \frac{1}{\sqrt{2\pi}}$ .

Part 2: Distribution of $\ln X$ (3 points)

*Note: Score exactly one path below | do not add across paths. This part tests the distribution transformation of a function of a random variable.*

Path A: Density function transformation method
Jacobian/derivative term [1 pt]: Write the inverse function $x=e^y$ of the transformation $y=\ln x$ and its derivative (or Jacobian determinant) $\frac{dx}{dy} = e^y$ .
Substitution and simplification [1 pt]: Correctly substitute $x=e^y$ and the derivative term 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 the complete standard normal probability density function (with domain $y\in\mathbb{R}$ ).
Path B: Distribution function method
Definition and transformation [1 pt]: Write the distribution function definition $F_Y(y) = P(\ln X \le y) = P(X \le e^y) = \int_{0}^{e^y} p(x) dx$ .
Integral substitution [1 pt]: Use the substitution $t=\ln x$ to transform the integration limits and integrand into the standard normal distribution function form $\int_{-\infty}^{y} c e^{-t^2/2} dt$ .
Final conclusion [1 pt]: Identify from the integral form that this is the standard normal distribution; conclusion same as Path A.
Shared prerequisite [max 1 pt]: If the student computed $c$ incorrectly in Part 1 but the derivation logic in this part is entirely correct, only the result point is deducted; process points are retained (follow-through).

Part 3: 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 / measure [1 pt]
State that the probability mass of the random vector $(X, \ln X)$ is concentrated on the plane curve $y = \ln x$ ; or state that its support has two-dimensional Lebesgue measure 0.
Determination and conclusion [1 pt]
Based on the above reasoning (measure 0 yet probability 1, or impossibility of writing a two-dimensional probability density $f(x,y)$ ), conclude: it is not a continuous random vector.

Total (max 7)

2. Zero-credit items

Copying the problem: Merely copying the given conditions or formula names (e.g., "use the density formula") without any concrete substitution or computation.
Unjustified guess in Part 3: In Part 3, merely answering "yes" or "no" without any mathematical argument.
Conceptual confusion: In Part 3, arguing "since the marginal distributions of $X$ and $\ln X$ are both continuous, the joint distribution is also continuous" (this conclusion is false; receives 0 points).
Incorrect independence assumption: In Part 3, attempting to construct a density via $f(x,y) = f_X(x) \cdot f_Y(y)$ (the problem does not give independence, and in fact the variables are perfectly dependent; this approach receives 0 points).

3. Deductions

*Apply at most one of the following (the most severe); total score cannot go below 0.*

Computation/arithmetic error (-1 pt):
Errors in constant handling during integration (e.g., missing $\sqrt{2\pi}$ , incorrect coefficients), leading to wrong values of $c$ or distribution parameters.
Missing domain (-1 pt):
When writing the probability density function as the final result, failing to specify the variable range (e.g., $y \in \mathbb{R}$ or $-\infty < y < +\infty$ ). However, this deduction is waived if the text explicitly states "follows the standard normal distribution."
Logical/notational confusion (-1 pt):
Severely confusing the random variable notation (uppercase $X$ ) with the value notation (lowercase $x$ ), or writing integration limits with confused logic (e.g., $0$ to $\ln x$ ), rendering the mathematical expression meaningless.

Probability Theory – Problem 64: Compute the value of ;