Probability Theory – Problem 69: Compute the value of ;

A 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.

1. Write the normalization condition and set up the integral The given density is $$p(x)=\dfrac{c}{x}e^{-\dfrac{(\ln x)^2}{2}},\quad x>0,$$ and 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 $dx=e^t dt$. 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 $+\infty$, so

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

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

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

3. Solve for the constant $c$

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

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

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

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

The transformation formula for a continuous random variable under a monotone invertible map 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 probability density function of the standard normal distribution.

3. State the conclusion

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

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

1. Recall the definition of a continuous (bivariate) random vector

A random vector $(U,V)$ is called a continuous random vector if there exists a nonnegative 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, for any set $B$ of planar Lebesgue measure zero, $$P\big((U,V)\in B\big)=0.$$

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 within 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 zero.

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 measure-zero set $A$ 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.

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.