The random variable follows an exponential distribution with parameter , and follows an exponential distribution with parameter . The two are mutually independent, with . 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 76: 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 $\lambda_{2}$ . The two are mutually independent, with $\lambda_{1}>\lambda_{2}>0$ . 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

1. CDF of the exponential distribution

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

2. Inverse CDF (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 CDF 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 $X=-\dfrac{1}{\lambda}\ln U$ .

3. Comparing inverse CDFs 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 $-\ln(1-u)>0$ . Given $\lambda_{1}>\lambda_{2}>0$ , it follows that $\dfrac{1}{\lambda_{1}}<\dfrac{1}{\lambda_{2}}.$

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

4. Choose a common uniform random variable

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

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

5. Construct the corresponding exponential random variables by parameter

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

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

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

By the same argument replacing $\lambda_{1}$ with $\lambda_{2}$ , we have $P(X_{2}'\le x)=1-e^{-\lambda_{2}x}$ , 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 $X_{2}$ .

6. Monotone comparison via the quantile function

By the conclusion from step 3, for any $u\in(0,1)$ , $F^{-1}_{\lambda_{1}}(u)\le F^{-1}_{\lambda_{2}}(u).$

In our construction, $U(\omega)\in(0,1)$ holds almost surely (the only exceptions are $U=0$ or $U=1$ , which have 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 terms, $P(X_{1}'\le X_{2}')=1.$

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, so overall $X_{1}'\le X_{2}'$ holds with probability $1$ , i.e., "almost surely".

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 marking scheme for this probability theory problem. The scheme maps the original 10-point problem to an integer score from 0 to 7.

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 for generating an exponential distribution from a uniform variable $U$ (e.g., $X = -\frac{1}{\lambda}\ln(1-U)$ or $-\frac{1}{\lambda}\ln U$ ). (2 pts)
[additive] Explicitly construct $X_1', X_2'$ , with the key point being the use of the same random variable $U$ (coupling) in both definitions. (2 pts)
[additive] Verify or explain that the constructed $X_1', X_2'$ respectively follow exponential distributions with parameters $\lambda_1, \lambda_2$ (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 the same sign, conclude $X_1' \le X_2'$ (almost surely). (2 pts)

Chain B: Direct Linear Transformation

[additive] Propose the idea based on quantile correspondence, i.e., set $F_1(X_1') = F_2(X_2')$ or let $X_1' \sim \text{Exp}(\lambda_1)$ and attempt to find $X_2' = g(X_1')$ . (2 pts)
[additive] Solve for 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 CDF proof). (1 pt)
[additive] Use the fact that the coefficient $\frac{\lambda_1}{\lambda_2} > 1$ and $X_1' \ge 0$ to directly conclude $X_2' \ge X_1'$ . (2 pts)

Total (max 7)

2. Zero-credit items

Only listing the PDF ( $f(x)$ ) or CDF ( $F(x)$ ) formula of the exponential distribution without performing inverse function computation or any construction attempt.
Constructing $X_1', X_2'$ independently (e.g., using independent $U_1, U_2$ ), which fails to ensure $X_1' \le X_2'$ 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 actual mathematical steps.

3. Deductions

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

Probability Theory – Problem 76: The random variable follows an exponential distribution with parameter , and follows an exponential distribution with…