TITLE: Distribution of parameter under the null hypothesis for mixture distributions
QUESTION [1 upvotes]: I am conducting a classical hypothesis test concerning the value of some parameter, i.e. $H_{0}:\theta=\theta_{0}$. I'll denote the  distribution of $\theta$ under the null as $f(\theta)$.
Suppose there is a 50% chance that $f(\theta)=g(\theta)$, a 30% chance $f(\theta)=h(\theta)$ and a 20% chance that $f(\theta)=m(\theta)$; where $g$, $h$ and $m$ are all known.
Intuitively, it doesn't seem to follow that one can use $0.5g(\theta)+0.3h(\theta)+0.2m(\theta)$ to form the distribution of $\theta$ under $H_{0}$ (the resulting distribution doesn't have a practical interpretation and of course won't consistently estimate the true distribution). Does anyone have any suggestions of what I could do? Is there even a solution to this problem? 
PS Not sure if relevant; but g, h and m were formed using a kernel nonparametric density estimator.
Edit:

The alternative hypothesis is $H_{A}:\theta≠\theta_{0}$.
$\theta$ is neither a scale nor location parameter. It is actually one of the coefficients (the intercept) in a LS-regression model, but I am applying an algorithm to get around a tricky (and rather unique) multiple comparisons problem. The final step in this is forming the null distribution to conduct a hypothesis test, and whilst I have calculated the vector of probabilities (0.5, 0.3, 0.2) and the corresponding null-distributions (g, h, m), I have became stuck with this final step!

REPLY [0 votes]: Comment:  I want to use simulation to demonstrate
the PDF of your mixture of distributions. Let the
first be $Exp(rate = 1)$ (50%), the second be $Norm(\mu=2,\sigma=1/2)$ (30%), and the third be $Unif(2,4)$ (20%). 
Simulation. The program below
makes these choices for 100,000 observations, plots the histogram,
and then plots the weighted average of the PDFs that you do not find
intuitive--and I do.
 B = 10^5;  x = k = numeric(m)
 for (i in 1:m) {
   h = sample(1:3, 1, prob=c(.5,.3,.2));  k[i] = h
   x[i] = (h==1)*rexp(1,1) + (h==2)*rnorm(1,2,.5) + (h==3)*runif(1,2,4) }

 table(k)/B
 ## k
 ##       1       2       3 
 ## 5.00413 2.99687 1.99900 

 mean(x);  sd(x)
 ## 1.699684   # apprx E(X)
 ## 1.118747   # apprx SD(X)

Comments on simulation results. It seems from table(k)/B that the distributions were chosen
in the intended proportions. Also, $E(X) = .5(1) + .3(2) + .2(3) = 1.7$ is well approximated. So the simulation program is working
as intended.
It is a little more difficult to find the variance $V(X)$ of the mixture distribution
because it reflects both the variances of the three individual
distributions and the scatter of their means. (See the Wikipedia article on 'mixture distributions' under moments.)
 hist(x, prob=T, col="skyblue2", ylim=c(0,.5))
 curve(.5*dexp(x,1)+.3*dnorm(x,2,.5)+.2*dunif(x,2,4), 
    lwd=2, col="blue", n=1001, add=T)


It seems that the weighted sum of PDFs is the correct PDF. I deliberately chose
three different distributional shapes in order to get a
combined distribution with an unusually shaped PDF and a relatively obvious
visual test of fit. (The bit of a tail to the right of 4 is
mainly due to the exponential component, partly to normal.)
Next? We will have to think about how to do the testing. Maybe something
like a one-sample t statistic will work, and we can discover
the critical value of its distribution through simulation. Because
you have seen some data, you may have better suggestions what to try next.
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