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I have this bivariate distribution and I would like to tell if it is in the exponential family: $$f(y_{i},d_{i}|\theta_{1}\,\theta_{2}\,p)=\left(\left(\frac{1}{\theta_{1}}\right)\exp\left(\frac{-y_{i}}{\theta_{1}}\right)\right)^{d_{i}}p^{d_{i}}\left(\left(\frac{1}{\theta_{2}}\right)\exp\left(\frac{-y_{i}}{\theta_{2}}\right)\right)^{1-d_{i}}(1-p)^{1-d_{i}}$$ with $p \in [0,1]; ~d_{i}\in\{0,1\}$ and $y_{i}\in(0,\infty).$

My claim is that this distribution is not a part of the exponential family because its support depends on the parameter $p.$ For example, if $p=0, $ then only pairs $(y_{i},0)$ are supported, but the solutions states that this distribution is an member of the exponential family. I am not sure where my misunderstanding is. Any explanation would be appreciated.

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    $\begingroup$ These expressions are neither valid densities nor distribution functions. You must be applying some implicit restrictions on the values of $y_i$ and $d_i.$ What are they? $\endgroup$
    – whuber
    Commented Sep 30, 2022 at 19:57
  • $\begingroup$ sorry d_{i} $\in$ {0,1} and $y_{i}\in\,(0,\infty)$ and it is a density function. $\endgroup$ Commented Sep 30, 2022 at 20:04
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    $\begingroup$ To be a valid density function, its integral over $(0,\infty)\times [0,1]$ must be $1$ for any possible values of the parameters $\theta_i$ and $p.$ That is not the case. The formula looks like some kind of corrupted version of a mixture model, but that's only a guess. $\endgroup$
    – whuber
    Commented Sep 30, 2022 at 22:09
  • $\begingroup$ d is binary taking values either 0 or 1. In which case I believe it would integrate to 1. When $\theta_{1}\in\,(0,\infty)$ and $\theta_{2}\in\,(0,\infty)$ $\endgroup$ Commented Sep 30, 2022 at 22:37
  • $\begingroup$ That change to your question helps. $\endgroup$
    – whuber
    Commented Oct 1, 2022 at 15:01

1 Answer 1

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As @whuber says, what you describe is a mixture model.

It looks like $Y \mid D=1 \sim \text{Exp}(\text{mean}=\theta_1)$ and $Y\mid D=0 \sim \text{Exp}(\text{mean}=\theta_2)$, where $D \sim \text{Bernoulli}(p)$.

In other words, $Y\sim \text{Exp}(\text{mean}=\theta_1)$ with probability $p$ and $Y\sim \text{Exp}(\text{mean}=\theta_2)$ with probability $1-p$.

For $\theta_1,\theta_2>0$ and $p\in [0,1]$, likelihood function given the data $(y,d)$ is

$$L(\theta_1,\theta_2,p)=\left\{p\frac1{\theta_1}e^{-y/\theta_1}\right\}^d\left\{(1-p)\frac1{\theta_2}e^{-y/\theta_2}\right\}^{1-d} \quad,\, y\ge 0,\,d\in\{0,1\}$$

Or,

$$L(\boldsymbol\theta)=\exp\left\{d\left(\ln \left(\frac p{\theta_1}\right)-\frac{y}{\theta_1}\right)+(1-d)\left(\ln \left(\frac {1-p}{\theta_2}\right)-\frac{y}{\theta_2}\right)\right\}$$

This can be written in the form

$$L(\boldsymbol\theta)=\exp\left\{\eta(\boldsymbol\theta)'T(y,d)-A(\boldsymbol\theta)\right\}\,,$$

where $\eta(\boldsymbol\theta)=\left(\ln\left(\frac{p\theta_2}{(1-p)\theta_1}\right),\frac1{\theta_2}-\frac1{\theta_1},-\frac1{\theta_2}\right)$, $T(y,d)=(d,dy,y)$ and $A(\boldsymbol\theta)=\ln\left(\frac{\theta_2}{1-p}\right)$.

Note that the support of the distribution is $\mathbb R_{\ge 0}\times \{0,1\}$, which does not depend on any parameter. So looks to me this is very much a member of exponential family.

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  • $\begingroup$ Thank you for your solution. Im still a bit confused why the support would not be a piecewise function of p. Like $\mathbb R_{\ge 0}\times \{0,1\}$ if p is not equal to 1 or 0, $\mathbb R_{\ge 0}\times \{1\}$ if p=1 and $\mathbb R_{\ge 0}\times \{0\}$ if p=0? $\endgroup$ Commented Oct 1, 2022 at 17:44
  • $\begingroup$ (+1) There's a (tiny) endpoint problem that I suspect is causing the concern: when $p$ is $0$ or $1,$ the support shrinks from a pair of rays to a single ray. It's just a slightly complicated version of the situation for, say, a Poisson family whose distribution degenerates when the parameter equals $0.$ The concept of a common reference measure ought to resolve this concern. $\endgroup$
    – whuber
    Commented Oct 1, 2022 at 18:03
  • $\begingroup$ @Iambereftoflordship The support does not change with $p$; the value of $d$ still remains $0$ or $1$. If you simply consider a Bernoulli$(p)$ model, then this is a member of exponential family as long as the parameter space contains an open subset of $\mathbb R$. And that is certainly the case if $p\in [0,1]$. $\endgroup$ Commented Oct 1, 2022 at 18:30
  • $\begingroup$ This makes sense. Thank you so much for your help. $\endgroup$ Commented Oct 1, 2022 at 20:53

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