I am stuck on how to solve this problem.

So, we have two sequences of random variables, $X_i$ and $Y_i$ for $i=1,...,n$. Now, $X$ and $Y$ are independent exponential distributions with parameters $\lambda$ and $\mu$. However, instead of observing $X$ and $Y$, we observe instead $Z$ and $W$.

$Z=\min(X_i,Y_i)$ and $W=1$ if $Z_i=X_i$ and 0 if $Z_i=Y_i$. I have to find closed-forms for the maximum likelihood estimators of $\lambda$ and $\mu$ on the basis of $Z$ and $W$. Further, we need to show that these are global maxima.

Now, I know that the minimum of two independent exponentials is itself exponential, with the rate equal to the sum of rates, so we know that $Z$ is exponential with parameter $\lambda+\mu$. Thus our maximum likelihood estimator is: $\hat{\lambda}+\hat{\mu}=\bar{Z}$.

But I'm stuck with where to go from here. I know that $W$ is a Bernoulli distribution with parameter $p=P(Z_i=X_i)$, but I don't know how to go about converting this into a statement about one of the parameters. For example, what would the MLE $\bar{W}$ be estimating in terms of $\lambda$ and/or $\mu$? I understand that if $Z_i=X_i$, then $\mu=0$, but I'm having a hard time figuring out how to come up with any algebraic statement, here.

UPDATE 1: So I have been told in the comments to derive the likelihood for the joint distribution of $Z$ and $W$.

So $f(Z,W)=f(Z|W=1)\cdot p+f(Z|W=0)\cdot (1-p)$ where $p=P(Z_i=X_i)$. Correct? I don't know how else to derive a joint distribution in this case, since $Z$ and $W$ are not independent.

So this gives us, $f(Z_i,W_i)=p\lambda e^{-\lambda z_i}+(1-p)\mu e^{-\mu z_i}$, by the definition of $W$ above. But now what? This doesn't get me anywhere. If I go through the steps of calculating the likelihood, I get: (using $m$ and $n$ as the sample sizes for each part of the mixture...)

$L(\lambda,\mu)=p^m\lambda^m e^{-\lambda \sum{z_i}}+(1-p)^n\mu^n e^{-\mu \sum{z_i}}$

$\log L=m\log p+m\log\lambda-\lambda \sum{z_i}+n\log(1-p)+n\log\mu-\mu \sum{z_i}$

If I take the partial derivatives, this tells me that my MLE estimates for $\lambda$ and $\mu$ are just the average of the $Z$'s conditional on $W$. That is,





  • 1
    $\begingroup$ Having just answered a similar MLE question today, may I direct you towards that solution for some ideas? The relationship between the questions is that your data also break up naturally into two disjoint groups: those where $W=0$ and those where $W=1$. It all comes down to writing down the likelihood for an observation of the form $(Z,W)=(z,0)$; the symmetry between $X$ and $Y$, $\mu$ and $\lambda$, immediately produces the likelihood for data of the form $(z,1)$ and then you're off and running. $\endgroup$
    – whuber
    Commented Nov 17, 2014 at 20:56
  • $\begingroup$ Do not rush to writing the maximum likelihood! First, express the joint distribution of $(Z,W)$, then deduce the likelihood associated with the sample of $(Z_i,W)=_i)$, which happens to be closed-form thanks to the exponential assumption. Then and only then you can try to maximise the function and hence derive the maximum likelihood. $\endgroup$
    – Xi'an
    Commented Nov 17, 2014 at 20:57
  • $\begingroup$ @whuber: (+1) it is rather straightforward indeed and involves the separation between the $(z_i,1)$'s and the $(z_i,0)$ but both groups involve both $\mu$ and $\lambda$, since they bring information on both $X_i$ and $Y_i$, since $W_i=\mathbb{I}(X_i<Y_i)$. $\endgroup$
    – Xi'an
    Commented Nov 17, 2014 at 21:00
  • 2
    $\begingroup$ @Xi'an That's right--and the parallels with the Normal-theory example I link to continue to hold, because there both groups provide information about the common parameter $\sigma$ (the scale), whose estimate will thereby involve "pooling" data from the groups. Here it will be seen that $\bar W$ tells us how the estimate of $\lambda+\mu$ (the rate, or inverse scale, for $Z$) should be apportioned into separate estimates of $\lambda$ and $\mu$. $\endgroup$
    – whuber
    Commented Nov 17, 2014 at 21:05
  • $\begingroup$ I've read through the other thread, whuber, but I honestly don't understand how to apply that to this example. Z and W aren't independent, so how do I derive the joint distribution? $\endgroup$ Commented Nov 17, 2014 at 22:04

1 Answer 1


I don't have enough points to comment, so I will write here. I think the problem you post can be viewed from a survival analysis perspective, if you consider the following:

$X_i$: True survival time,

$Y_i$: Censoring time,

Both have an exponential distribution with $X$ and $Y$ independent. Then $Z_i$ is the observed survival time and $W_i$ the censoring indicator.

If you are familiar with survival analysis, I believe you can start from this point.

Notes: A good source: Analysis of Survival Data by D.R.Cox and D.Oakes

Below is an example: Assuming the p.d.f of the survival time distribution is $f(t)=\rho e^{-\rho t}$. Then the survival function is: $S(t)=e^{-\rho t}$. And the log-likelihood is:

$\mathcal{l}= \sum_u \log f(z_i) + \sum_c \log S(z_i)$

with summation over uncensored people ($u$) and censored people ($c$) respectively.

Due to the fact that $f(t)=h(t)S(t)$ where h(t) is the hazard function, this can be written:

$\mathcal{l}= \sum_u \log h(z_i) + \sum \log S(z_i)$

$\mathcal{l}= \sum_u \log \rho - \rho \sum z_i$

And the maximum likelihood estimator $\hat{\rho}$ of $\rho$ is:

$\hat{\rho}=d/\sum z_i$ where $d$ is the total number of cases of $W_i=1$


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