The random variable follows an exponential distribution with parameter , and follows an exponential distribution with parameter , where and are independent and . Construct a random variable having the same distribution as and a random variable having the same distribution as such that almost surely.

Probability Theory Solution – Problem 18: The random variable follows an exponential distribution with parameter , and follows an exponential distribution with…

Question

The random variable $X_{1}$ follows an exponential distribution with parameter $\lambda_{1}$ , and $X_{2}$ follows an exponential distribution with parameter $\lambda_{2}$ , where $X_{1}$ and $X_{2}$ are independent and $\lambda_{1}>\lambda_{2}>0$ . Construct a random variable $X_{1}^{\prime}$ having the same distribution as $X_{1}$ and a random variable $X_{2}^{\prime}$ having the same distribution as $X_{2}$ such that $X_{1}^{\prime}\leqslant X_{2}^{\prime}$ almost surely.

Step-by-step solution

Step 1. Distribution function of the exponential distribution

Let the random variable $X$ follow an exponential distribution with parameter $\lambda>0$ . Its probability density function is $f(x)=\lambda e^{-\lambda x}$ ( $x>0$ ), and hence its distribution function is $F(x)=P(X\le x)= \begin{cases} 0,\quad x\le 0\\ 1-e^{-\lambda x},\quad x>0 \end{cases}$ .

Step 2. Inverse distribution function (quantile function)

For $0<u<1$ , set $F(x)=u$ , i.e., $1-e^{-\lambda x}=u \Longrightarrow e^{-\lambda x}=1-u \Longrightarrow -\lambda x=\ln(1-u) \Longrightarrow x=-\dfrac{1}{\lambda}\ln(1-u).$

Therefore the inverse distribution function is $F^{-1}(u)=-\dfrac{1}{\lambda}\ln(1-u),\quad 0<u<1.$ If $U\sim \text{Uniform}(0,1)$ , setting $X=F^{-1}(U)=-\dfrac{1}{\lambda}\ln(1-U)$ yields $X$ following an exponential distribution with parameter $\lambda$ . Note that $1-U$ has the same distribution as $U$ , so one may also write $X=-\dfrac{1}{\lambda}\ln U$ .

Step 3. Comparison of inverse distribution functions for different parameters

For the same $u\in(0,1)$ , consider $F^{-1}_{\lambda_{1}}(u)=-\dfrac{1}{\lambda_{1}}\ln(1-u),\quad F^{-1}_{\lambda_{2}}(u)=-\dfrac{1}{\lambda_{2}}\ln(1-u).$

Since $0<u<1$ , we have $1-u\in(0,1)$ , so $\ln(1-u)<0$ , and thus $-\ln(1-u)>0$ . Moreover, since $\lambda_{1}>\lambda_{2}>0$ , it follows that $\dfrac{1}{\lambda_{1}}<\dfrac{1}{\lambda_{2}}.$

Multiplying both sides by the positive quantity $-\ln(1-u)$ gives $-\dfrac{1}{\lambda_{1}}\ln(1-u)\le -\dfrac{1}{\lambda_{2}}\ln(1-u),$ that is, $F^{-1}_{\lambda_{1}}(u)\le F^{-1}_{\lambda_{2}}(u),\quad 0<u<1.$

Step 4. Selection of a common uniform random variable

On a suitable probability space, let $U$ be a random variable satisfying $U\sim \text{Uniform}(0,1).$

We shall use $U$ alone to construct the new random variables $X_{1}'$ and $X_{2}'$ .

Step 5. Construction of the corresponding exponential random variables

Define $X_{1}'=-\dfrac{1}{\lambda_{1}}\ln(1-U),\quad X_{2}'=-\dfrac{1}{\lambda_{2}}\ln(1-U).$

For any $x>0$ , we have $P(X_{1}'\le x) =P\left(-\dfrac{1}{\lambda_{1}}\ln(1-U)\le x\right) =P\left(1-U\ge e^{-\lambda_{1}x}\right)$ .

Simplifying further: $1-U\ge e^{-\lambda_{1}x} \Longleftrightarrow U\le 1-e^{-\lambda_{1}x}$ . Since $U\sim \text{Uniform}(0,1)$ , we have $P(U\le t)=t$ ( $0<t<1$ ), and therefore $P(X_{1}'\le x) =1-e^{-\lambda_{1}x}.$ This coincides exactly with the distribution function of the exponential distribution with parameter $\lambda_{1}$ , so $X_{1}'$ follows an exponential distribution with parameter $\lambda_{1}$ .

By an analogous argument, $P(X_{2}'\le x)=1-e^{-\lambda_{2}x}$ , i.e., $X_{2}'$ follows an exponential distribution with parameter $\lambda_{2}$ .

Therefore, $X_{1}'$ has the same distribution as $X_{1}$ , and $X_{2}'$ has the same distribution as $X_{2}$ .

Step 6. Monotone comparison via quantile functions

From the conclusion in Step 3, for every $u\in(0,1)$ we have $F^{-1}_{\lambda_{1}}(u)\le F^{-1}_{\lambda_{2}}(u).$

By construction, $U(\omega)\in(0,1)$ holds almost surely (the only exceptions are $U=0$ and $U=1$ , each of which has probability $0$ ), and $X_{1}'(\omega)=F^{-1}_{\lambda_{1}}(U(\omega)),\quad X_{2}'(\omega)=F^{-1}_{\lambda_{2}}(U(\omega)).$

Therefore, for almost every $\omega$ , $X_{1}'(\omega)\le X_{2}'(\omega).$

In probabilistic notation, $P(X_{1}'\le X_{2}')=1.$

Step 7. Explanation of "almost surely"

Since $U$ is a continuous random variable, $P(U=0)=0,\quad P(U=1)=0,$ so $P(U\in(0,1))=1$ . When $U\in(0,1)$ , the above inequality holds strictly. Therefore $X_{1}'\le X_{2}'$ holds with probability $1$ , i.e., almost surely. QED.

Final answer

Let $U\sim \text{Uniform}(0,1)$ and define $X_{1}'=-\dfrac{1}{\lambda_{1}}\ln(1-U),\quad X_{2}'=-\dfrac{1}{\lambda_{2}}\ln(1-U).$

Marking scheme

The following is the rubric for this probability theory problem. This rubric assumes the official solution is fully correct and maps the original 10-point problem to an integer scoring scale of 0 to 7 points.

1. Checkpoints (max 7 pts total)

Score exactly one chain | take the maximum subtotal among chains; do not add points across chains.

Chain A: Inverse Transform Method (Official Approach)

[additive] Derive the inverse CDF or directly cite the result that a uniform random variable $U$ generates an exponential random variable via $X = -\frac{1}{\lambda}\ln(1-U)$ or $-\frac{1}{\lambda}\ln U$ . (2 pts)
[additive] Explicitly construct $X_1', X_2'$ using the same random variable $U$ (coupling). (2 pts)
[additive] Verify or explain that the constructed $X_1', X_2'$ follow exponential distributions with parameters $\lambda_1, \lambda_2$ respectively (correctness of marginal distributions). (1 pt)
[additive] Use $\lambda_1 > \lambda_2$ to derive the inequality: since $1/\lambda_1 < 1/\lambda_2$ and the logarithmic term has a consistent sign, conclude that $X_1' \le X_2'$ almost surely. (2 pts)

Chain B: Direct Linear Transformation

[additive] Propose a quantile-matching approach, i.e., set $F_1(X_1') = F_2(X_2')$ or let $X_1' \sim \text{Exp}(\lambda_1)$ and seek $X_2' = g(X_1')$ . (2 pts)
[additive] Derive the explicit linear relation $X_2' = \frac{\lambda_1}{\lambda_2} X_1'$ . (2 pts)
[additive] Verify that when $X_1' \sim \text{Exp}(\lambda_1)$ , the transformed $X_2'$ indeed follows $\text{Exp}(\lambda_2)$ (via change of variables or distribution function argument). (1 pt)
[additive] Use the fact that $\frac{\lambda_1}{\lambda_2} > 1$ together with $X_1' \ge 0$ to directly conclude $X_2' \ge X_1'$ . (2 pts)

Total (max 7)

2. Zero-credit items

Merely listing the PDF ( $f(x)$ ) or CDF ( $F(x)$ ) of the exponential distribution without computing the inverse function or attempting a construction.
Constructing $X_1', X_2'$ independently (e.g., using independent $U_1, U_2$ ), which fails to guarantee $X_1' \le X_2'$ almost surely (the probability would only be $P(X_1 \le X_2) < 1$ ).
Merely restating the problem requirements ("we need to construct...") without any substantive mathematical steps.

3. Deductions

Logic Gap: Failing to define the auxiliary random variable (e.g., not stating $U \sim U(0,1)$ or its range), while the derivation implicitly reflects correct understanding. (-1)
Logic Gap: Construction is correct in form but does not mention or verify "almost surely" ( $P(\cdot)=1$ ), or overlooks the minor rigor issue that $\ln(1-U)$ is undefined at $U=1$ (in undergraduate grading, if the overall logic is sound, this is typically not penalized). (No Penalty)
Fatal Error: Attempting an additive construction $X_2' = X_1' + Y$ (where $Y \ge 0$ and independent), which causes $X_2'$ to no longer follow an exponential distribution (convolution destroys the distributional family); this constitutes a fundamental error. (Cap score at 3/7 for attempting a construction, unless a rigorous proof of the resulting distribution is provided)

Probability Theory – Problem 18: The random variable follows an exponential distribution with parameter , and follows an exponential distribution with…