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Why is it true that $\gamma(0)=1$? By saying that $\gamma(0)=1$, we are saying that $\mathbb{P}_0\left(f_1\left(X\right) > 0\right) =1$. Since $f_1$ is non-negative, we are saying that $\mathbb{P}_0\left(f_1\left(X\right) = 0 \right) = 0$. To me, this is not obviously true.

To clarify things, the author use the term p.d.f. to refer to the Radon-Nikodym derivative $\frac{d P_i}{d v}$. The probability spaceis taken to be $\left(\mathbb{R}^n, \mathcal{B}^n\right)$, equipped with three measure: $\mathbb{P}_0$, $\mathbb{P}_1$ and $v$. In case where dominating measure $v$ is Lebesgue, $\frac{d \mathbb{P}_i}{dv}: \mathbb{R}^n \to \mathbb{R}$ is just the (joint) probability density function.

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Why is it true that $\gamma(0)=1$?

Just plug in $0$ for $t$: $$ \gamma(0) = P_0(f_1(X) > 0) = 1. $$ $f_1$ is a pdf. It has to evaluate to something positive. Well, as long as your're not plugging in data $X$ that are outside of $f_1$'s support.

Edit: regarding @whuber's comments, if they don't share the same support, I guess it isn't true, and there is a case where $\gamma(0) < 1$. Consider a shifted exponential distribution $f(x) = (x-\theta)e^{-(x-\theta)}$. If $f_0(x) = e^{-(x-1)}$ and $f_1(x) = e^{-(x-2)}$, then $P_0(f_1(X) = 0) = P_0(1<X < 2) > 0$.

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    $\begingroup$ You implicitly assume $f_1$ is continuous with respect to $f_0.$ Although that might be one of the assumptions in the reference, it doesn't explicitly appear anywhere in this thread. Why couldn't it be the case that $f_1$ assigns zero probability density to an event that has positive probability under $f_0$? $\endgroup$
    – whuber
    Commented May 6, 2018 at 19:13
  • $\begingroup$ No. $f_1$ is not necessarily a pdf (although if it is continuous pdf in the usual sense, the proof is less technical). All we know is that $f_1$ is the RN-derivative $\frac{d P_0}{d v}$. We know that $f_1$ is non-negative. And yes, I did make a typo. $\endgroup$
    – 3x89g2
    Commented May 6, 2018 at 19:17
  • $\begingroup$ @whuber good call. See edit. $\endgroup$
    – Taylor
    Commented May 6, 2018 at 19:34
  • $\begingroup$ @3x89g2 they do say pdf, but I see your point that $\nu$ isn't necessarily Lebesgue. I have just included an example supporting your initial belief; however, in that case, $\nu$ is Lebesgue. Also, I believe you mean $f_1$ is $\frac{d P_1}{d\nu}$. $\endgroup$
    – Taylor
    Commented May 6, 2018 at 19:36
  • $\begingroup$ Yes. Another typo. And I'm not sure if I'm following. Why is $f_0\left(x\right) = e^{-(x-1)}$ a density? It doesn't integrate to $1$. $\endgroup$
    – 3x89g2
    Commented May 6, 2018 at 19:40

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