# Does this distribution belong to exponential family and does its support depend on $p?$

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.

• 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?
– whuber
Commented Sep 30, 2022 at 19:57
• sorry d_{i} $\in$ {0,1} and $y_{i}\in\,(0,\infty)$ and it is a density function. Commented Sep 30, 2022 at 20:04
• 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.
– whuber
Commented Sep 30, 2022 at 22:09
• 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 $\theta_{2}\in\,(0,\infty)$ Commented Sep 30, 2022 at 22:37
• That change to your question helps.
– whuber
Commented Oct 1, 2022 at 15:01

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.

• 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? Commented Oct 1, 2022 at 17:44
• (+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.
– whuber
Commented Oct 1, 2022 at 18:03
• @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 $p\in [0,1]$. Commented Oct 1, 2022 at 18:30
• This makes sense. Thank you so much for your help. Commented Oct 1, 2022 at 20:53