Support of likelihood ratio test statistic Say I'm testing $H_0: Y \sim \text{Exp}(1)$ against $H_1: Y \sim \text{U}(0, 1)$. I believe this gives me the following likelihood ratio test:
$$
t^*(y) = \frac{p_1(y)}{p_0(y)}
 = \frac{1}{e ^ {-y}}
 = e ^ {y}
$$
The problem is defining the support for this statistic. Since $p_0(y)$ is defined $\forall y > 0$ and $p_1(y)$ $\forall y \in (0, 1)$, I don't know what to make of the support of the function $t^*(y)$, could I simply make it the intersection between $y > 0$ and $0 < y < 1$ (i.e., $0 < y < 1$)?
 A: This statistic weighs evidence for the two hypotheses by comparing their probability densities at the observed value of $y$.  Because the denominator could be zero in this situation, we have to consider two possibilities:


*

*The denominator is positive.  This means that $H_0$ assigns a positive chance to any tiny neighborhood of $y$.  It occurs when $0 \lt y$.  There is no problem with a division by zero.  In terms of the indicator function $\mathcal{I}$, a formula for the likelihood ratio is $$\frac{\mathcal{I}_{(0,1)}(y)}{e^{-y}}.$$  This equals $e^y$ for $0 \lt y \lt 1$ and otherwise is zero.

*The denominator is zero.  This means $H_0$ assigns no probability density to $y$.  There are two possibilities:


*

*$H_1$ assigns no probability density to $y$, either.  Thus, this $y$ has no chance of being observed under either hypothesis.  We needn't consider this any further.  The set of $y$ for which this is the case is the intersection of the complements of the supports of the hypotheses: the non-positive real numbers.

*$H_1$ assigns some probability density to $y$.  Thus, this $y$ is possible under $H_1$ but not under $H_0$.  The conclusion is obvious.  As a convention we may use values in an extended Real number line $\{-\infty, \infty\}\cup \mathbb{R}$ to designate such likelihood ratios (or their logarithms); here we would say that the likelihood ratio (and its log) is $\infty$.
To summarize, let $S_i\subset\mathbb{R}$ be the supports of the hypotheses.  Then the likelihood ratio must be considered a function whose domain is the union of supports $S_0\cup S_1$ which takes values in the extended positive reals $[0,\infty)\cup\{\infty\}$.  The log likelihood ratio takes values in the extended reals $\mathbb{R}\cup\{-\infty,\infty\}$, with $\log(0)$ defined to be $-\infty$.
When $S_0=S_1$, the zero-denominator case has no chance of happening, regardless of the hypothesis, and we may dispense with using the extended reals if we wish.  This is a frequent assumption in likelihood ratio settings.
A: Actually both densities are defined over the whole line, they're just 0 elsewhere than the places you mention.
You have to think carefully about the density across at least the +ve half-line -- your ansẃer defines what you get when $0<y<1$, but what's the LR when y=4.3? 
That could happen, if the distribution really were exponential, so you have to consider it.
