Checking whether a density is exponential family Trying to prove that this doesn't belong to exponential family.
$f(y|a)=4\frac{(y+a)}{(1+4a)} ;  0 < y < 1 , a>0$
Here is my approach:
$$f(y|a) = 4(y+a)e^{-log(1+4a)}$$
$$f(y|a) = (4y)(1+\frac{a}{y})e^{-log(1+4a)}$$
Comparing it with standard form, $h(y) = 4y$ and $g(a)$ which has to be a function of only $a$, can not be defined in terms of $a$ alone, as $y$ in the $1+\frac{a}{y}$ is inseparable. Is this enough to show that this distribution doesn't belong to exponential family.
Please review my approach.
 A: You've put your finger on the crux of the matter, and indeed the result is fairly obvious, but the logic seems a little off.  The method described below repeatedly uses logarithms and differentiation to make the problem progressively simpler, until it becomes utterly trivial.

By definition, $f$ is the PDF for an exponential family when its logarithm can be written as a sum of something in terms of the parameter ($a$) only, something else in terms of the data ($y$) only, and something else that is a product of a function of $a$ and a function of $y$.  This means you are free to ignore any factors that clearly depend only on the parameter or only on the data.  In this case it's obvious $\frac{4}{1+4a}$ depends only on $a$, so we may ignore it.  
The issue is with $y+a$.  We need to prove there cannot exist "nice" functions $\eta$ and $T$ such that $\log(y+a) = \eta(a) T(y)$ plus some function of $a$ alone plus some other function of $y$ alone.  That "plus" part is annoying, but we can kill it off by differentiating first with respect to $a$ (the derivative of any function of $y$ alone will be zero) and then with respect to $y$ (the derivative of a function of $a$ alone will be zero).  Upon negating both sides (to make the left hand side positive) this gives
$$-\frac{\partial^2}{\partial a\partial y}\log(y+a) = \frac{1}{(a+y)^2} = -\eta'(a)T'(y).$$
I want to take logarithms to simplify the right hand side (which, being equal to the left hand side, is always positive).  Assuming $\eta'$ and $T'$ are both continuous will guarantee there are some intervals of values for $a$ and for $y$ in which either $-\eta'(a)\gt 0$ and $T'(y)\gt 0$ or else $\eta'(a)\gt 0$ and $-T'(y)\gt 0$.  This means we indeed can split the right hand side into two positive factors, allowing the logarithm to be applied.  Doing so yields
$$-2\log(a+y) = \log\frac{1}{(a+y)^2} =  \log\left(-\eta'(a)T'(y)\right) = \log(-\eta'(a)) + \log(T'(y))$$
(or else a comparable expression with a few minus signs thrown in).  Now we play the same game: in either case, differentiating both sides with respect to both $a$ and $y$ yields
$$\frac{2}{(a+y)^2} = 0,$$
an impossibility.
Looking back, this approach had to assume both $\eta$ and $T$ have second derivatives within some intervals of their arguments.  The analysis can be done along the same lines using finite differences to weaken those assumptions, but it's probably not worth the bother.
A: It will be in exponential family if it can be written in $fh(x)e^{ηT(x)-A(η)}$(with another conditions ) . 

Let $g(x,η)=ηT(x)-A(η)$. Now for any 4 data points $x_1,x_2,x_3,x_4$ in the sample space
$ \frac{(g(x1,η)-g(x2,η))}{(g(x3,η)-g(x4,η))}$= $\frac{(T(x1)-T(x2))}{(T(x3) - T(x4))}$ , which is free of η.

Now here $f(y;a)= 4\frac{y+a}{(4a+1)}$ = $4 e^( ln(y+a)-ln(4a+1) )$.
Take $g(x,η)=ln(y+a)-ln(4a+1)$.

So
$\frac{(g(x1,η)-g(x2,η))}{(g(x3,η)-g(x4,η))}$=$\frac{(ln(y1+a)-ln(y2+a))}{(ln(y3+a)-ln(y4+a))}$
- which is not free of a .
so $f(y;a)$ does not belong to Exponential family .
