# Find UMVUE of difference of parameters of two exponential distribution random variables

Let $$X_{1}, \dots, X_{n}$$ be i.i.d. having the exponential distribution $$Exp\left(0, \theta_{x}\right)$$ with $$\theta_{x}>0$$, and $$Y_{1}, \dots, Y_{n}$$ be i.i.d. having the exponential distribution $$Exp\left(0,\theta_{y}\right)$$ with $$\theta_{y}>0$$. Assume that $$X_{i}$$'s and $$Y_{j}$$'s are independent, but they are unobservable.

Suppose that our sample is $$\left(Z_{1}, \Delta_{1}\right), \dots,\left(Z_{n}, \Delta_{n}\right)$$, where $$Z_{i}= X_{i}(1-\Delta_{i}) + Y_{i}\Delta_{i}$$ and $$\Delta_{i}=I(X_i \ge Y_i)$$ for $$i=1, \dots, n$$. Please find an unbiased estimator and the UMVUE of $$\theta_{x}-\theta_{y}$$.

Note that the probability density function (p.d.f.) of $$Exp(a,\theta)$$ is $$\theta^{-1}e^{-(x-a)/\theta}I(x>a)$$.

The parameter, $$\theta$$, is commonly called the MTBF (mean time between failures) or MTTF (mean time to fix). The inverse of the parameter, $$1/\theta$$, is equal to the hazard rate.

This question is a part of Exercise 3.9 in Shao Jun (2003).

The following is my attempt.

Here is the skeleton of my solution. We first find the complete and sufficient statistics $$T$$, and then try to construct the unbiased estimator $$\widehat{S}$$, then the UMVUE would be $$\mathbb{E}[\widehat{S}\vert T]$$.

Let $$U_i = X_i - Y_i$$. When $$u\le 0$$, we have \begin{align*} P(U_i \le u) & = P(X_i- Y_i \le u) \\ & = \int_{0}^{\infty} \int_{x-u}^{\infty} \frac{1}{\theta_x\theta_y} \exp\left\{ -\frac{x}{\theta_x} -\frac{y}{\theta_y} \right\} d y d x \\ & = \exp (u / \theta_y) \frac{\theta_y}{\theta_x + \theta_y} . \end{align*} This implies $$P(\Delta_i = 1) = \frac{\theta_x}{\theta_x + \theta_y}$$. Further, \begin{align*} P ( Z_i \le z, \Delta_i = 1 ) & = \frac{\theta_x}{\theta_x + \theta_y} - \frac{\theta_x}{\theta_x + \theta_y} \exp \left( - \frac{\theta_x + \theta_y}{\theta_x \theta_y} z\right) \triangleq P_{1,i} , \\ % & = P ( Z_i \le z, \Delta_i = 0 ) & = \frac{\theta_y}{\theta_x + \theta_y} - \frac{\theta_y}{\theta_x + \theta_y} \exp \left( - \frac{\theta_x + \theta_y}{\theta_x \theta_y} z\right) \triangleq P_{0,i} . \end{align*} Then $$Z_i$$ and $$\Delta_i$$ are independent, $$Z_i\sim E\left(0, \frac{\theta_x\theta_y}{\theta_x+\theta_y}\right)$$, and $$P_{Z_i,\Delta_i} (z,\delta) = P_{1,i}^{\delta} P_{0,i}^{1-\delta}$$.

Notice that the p.d.f. of the joint distribution of $$Z_i$$'s and $$\Delta_i$$'s is given by \begin{align*} & p\left( Z_1 = z_1, \dots, Z_n = z_n, \Delta_1 = \delta_1, \dots, \Delta_n = \delta_n \right) \\ & = \prod_{i=1}^n \left[ \frac{\theta_x + \theta_y}{\theta_x \theta_y} \exp\left( - \frac{\theta_x + \theta_y}{\theta_x \theta_y} z_i \right) \left\{ \frac{\theta_x}{\theta_x + \theta_y} I (\delta_i = 1) + \frac{\theta_y}{\theta_x + \theta_y} I (\delta_i = 0)\right\} \right] I(z_i>0)\\ & = \left(\frac{\theta_x + \theta_y}{\theta_x \theta_y}\right)^{n} \exp\left( - \frac{\theta_x + \theta_y}{\theta_x \theta_y} \sum_{i=1}^n z_i \right) \left( \frac{\theta_x}{\theta_x + \theta_y} \right)^{\sum_{i=1}^n \delta_i} \left( \frac{\theta_y}{\theta_x + \theta_y} \right)^{n- \sum_{i=1}^n \delta_i} I(z_{(1)}>0) \\ & = \exp\left( - \frac{\theta_x + \theta_y}{\theta_x \theta_y} \sum_{i=1}^n z_i \right) \theta_x^{-n+\sum_{i=1}^n \delta_i} \theta_y^{- \sum_{i=1}^n \delta_i} I(z_{(1)}>0) \end{align*} Therefore, the joint distribution of $$Z_i$$'s and $$\Delta_i$$'s is from an exponential family with $$T = (\sum_{i=1}^n \Delta_i, \sum_{i=1}^n Z_i)$$ as the complete and sufficient statistic.

I don't know how to continue and if it will actually exist. Thanks for any suggestions and answers.

(update)

Thanks for the suggestion from @Xi'an.

If we are interested in $$1/\theta_x - 1/\theta_y$$, we could construct an unbiased estimator according to $$\frac{n-1}{n} \mathbb{E} \frac{\sum_{i=1}^n (1-\Delta_i)}{\sum_{i=1} Z_i} - \frac{n-1}{n} \mathbb{E} \frac{\sum_{i=1}^n \Delta_i}{\sum_{i=1} Z_i} = 1/\theta_x - 1/\theta_y$$.

But in this question, we focus on $$\theta_x -\theta_y$$ Although $$\mathbb{E}[X_i] = \theta_x$$, we have $$\mathbb{E}[X_i \vert Z_i,\Delta_i]$$ is a function of $$\theta_x$$, $$Z_i$$ and $$\Delta_i$$. However, $$\mathbb{E}[X_i \vert Z_i,\Delta_i]$$ is non-linear in terms of $$\theta_x$$. So, I don't know how to continue.

• In effect, you observe an iid sample $(Z_i)$ of an exponential distribution with (rate) parameter $1/\theta=1/\theta_x+1/\theta_y.$ Since $\theta$ is a symmetric function of the parameters, if you could find an unbiased estimator of $\theta_x-\theta_y$ it would necessarily also be an unbiased estimator of $\theta_y-\theta_x,$ immediately implying $\theta_x=\theta_y.$ Since you don't assume equality of the parameters, your objective is hopeless: such an unbiased estimator doesn't exist.
– whuber
Mar 20, 2022 at 13:59
• @whuber Thanks a lot! Yes, parameter of $Z$, i.e., $\theta$ is a symmetric function of the parameters $\theta_x$ and $\theta_y$. But, I think, it may not be sufficient to show that the unbiased estimator doesn't exist. Because, $\Delta$ contains the information of $I(\theta_x \ge \theta_y)$. Mar 20, 2022 at 14:18
• Thank you--I thought I might be overlooking something. So that means you should be able to use the mean of the $\Delta_i$ to estimate the chance with which one distribution will exceed the other--and that ought to lead to an effective strategy for estimating another function of $(\theta_x,\theta_y)$ from which you can obtain estimators of both parameters.
– whuber
Mar 20, 2022 at 14:26
• The fascinating thing is that they are not symmetrical, as shown by my "answer". Mar 22, 2022 at 7:21
• @Xi'an Yes. This question is a part of Exercise 3.9 in Shao Jun (2003). I added the source in my question. Mar 22, 2022 at 11:09

1. If $$\theta_x$$ and $$\theta_y$$ are rate rather than scale parameters, $$\frac{n-1}{n} \frac{\sum_{i=1}^n (1-\Delta_i)}{\sum_{i=1} Z_i} - \frac{n-1}{n} \frac{\sum_{i=1}^n \Delta_i}{\sum_{i=1} Z_i}\tag{1}$$ is an unbiased estimator of $$1/\theta_x - 1/\theta_y$$ and since it only depends on $$\mathbf T$$, it is the UMVUE.
2. If instead $$Z=\max\{X,Y\}$$, with $$\theta_x$$ and $$\theta_y$$ scale parameters, the conditional distribution of $$Z$$ conditional on $$\Delta=1$$. Since $$\mathbb P(\Delta=1)=\frac{\theta_x}{\theta_x+\theta_y}$$ we have \begin{align} \mathbb P(Z\le z|\Delta=1)&=\frac{\theta_x+\theta_y}{\theta_x}\mathbb P(Z\le z,\Delta=1)\\ &=\frac{\theta_x+\theta_y}{\theta_x}\mathbb P(X\le z,X>Y)\\ &=\frac{\theta_x+\theta_y}{\theta_x}\int_0^z\int_0^x \frac1{\theta_x\theta_y} \exp\{-x/\theta_x-y/\theta_y\}\text dx\text dy\\ &=\frac{\theta_x+\theta_y}{\theta_x}\int_0^z(1-\exp\{-x/\theta_y\})\frac{\exp\{-x/\theta_x\}}{\theta_x}\text dx\\ &=\frac{\theta_x+\theta_y}{\theta_x}[1-\exp\{-z/\theta_x\}]-\\ &\qquad\frac{\theta_x+\theta_y}{\theta_x}(\theta_x^{-1}+\theta_y^{-1})^{-1} [1-\exp\{-z(\theta_x^{-1}+\theta_y^{-1})\}]\\ &=\frac{\theta_x+\theta_y}{\theta_x}[1-\exp\{-z/\theta_x\}]- \frac{\theta_y}{\theta_x}[1-\exp\{-z(\theta_x+\theta_y)/\theta_x\theta_y\}] \end{align} This is a signed mixture of two Exponential distributions $$\frac{\theta_x+\theta_y}{\theta_x}\mathcal Exp(\theta_x)-\frac{\theta_y}{\theta_x}\mathcal Exp(\theta_x\theta_y/(\theta_x+\theta_y))$$ which is illustrated by the fit in the following graphs:
based on $$n=10^6$$ simulations from $$\mathcal Exp(10)$$ and $$\mathcal Exp(1/10)$$ samples. This distribution has mean \begin{align}\mathbb E[Z\le z|\Delta=1] &=\frac{\theta_x+\theta_y}{\theta_x}\theta_x-\frac{\theta_y}{\theta_x}\frac{\theta_x\theta_y}{\theta_x+\theta_y}\\ &=\theta_x+\theta_y\left[1-\frac{\theta_y}{\theta_x+\theta_y}\right]\\ &=\theta_x+\frac{\theta_y\theta_y}{\theta_x+\theta_y}\end{align} The second term above is symmetric in $$(\theta_x,\theta_y)$$. Therefore, $$\mathbb E[Z\le z|\Delta=1]-\mathbb E[Z\le z|\Delta=0]=\theta_x-\theta_y$$ which leads immediately to an unbiased estimator based on $$(\mathbf X,\boldsymbol \Delta)$$: $$\dfrac{\sum_{i=1}^n Z_i\Delta_i}{\sum_{i=1}^n\Delta_i}- \dfrac{\sum_{i=1}^n Z_i\{1-\Delta_i\}}{\sum_{i=1}^n\{1-\Delta_i\}}\tag{2}$$ although I cannot tell about about (2) being UMVUE as the $$Z_i$$'s are not from an exponential family.