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I am to prove that:

1) The expected value of $(Λ^n|H_1) = E(Λ^{(n+1)}|H_0)$

2) The expected value of $(Λ|H_0) = 1 $

where $Λ$ is the likelihood ratio. I know that the likelihood ratio is equal to $f(x|H_1)/f(x|H_0)$ but my textbook does not mention any relationships between the likelihood ratio to expected values. I am still fairly new to these statistics topics so any help/hints is appreciated. Thanks in advance!

Jarle Tufto
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Raptors1102
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    Hint for #2: the likelihood ratio $\displaystyle \Lambda(X) = \frac{f_1(X)}{f_0(X)}$ is a function of the random variable $X$. Conditioned on $H_0$ being the true hypothesis, $X$ has density $f_0(X)$. Do you know how to find the expected value of a function $\Lambda(X)$ of $X$ directly without first determining the density of $\Lambda(X)$, that is, just from knowledge of the density function $f_0(x)$? – Dilip Sarwate Feb 16 '15 at 21:11
  • @DilipSarwate Would you use the intergral of x*f(x) dx? – Raptors1102 Feb 16 '15 at 21:44
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    $\displaystyle \int_{-\infty}^{\infty} xf(x),\mathrm dx$ will give you the expected value of $X$, not of $\Lambda(X)$. What you need to know about finding $E[\Lambda(X)]$ is perhaps the very next equation in your book. – Dilip Sarwate Feb 16 '15 at 22:13
  • @DilipSarwate Ok, I think I might be on the right track. Would I use E[Λ(X)] = intergral of Λ(x)f(x) dx? The inner product of Λ and f. Thanks for your time by the way – Raptors1102 Feb 16 '15 at 22:54
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    Yes, you are on the right track. Just remember that when you find $E[\Lambda(X)\mid H_0]$ you need to use $f_0(x)$ as the density for $X$. – Dilip Sarwate Feb 16 '15 at 23:02
  • So this is easy when H0 is the true hypothesis, but what about when H1 is? Given how the H0 case works out so nicely I thought the answer for H1 should also be easy, but I'm not seeing it. – Ben Farmer Aug 24 '17 at 18:36

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Now that any homework deadlines are long past, assuming a continuous random variable $X$ whose densities $f_0(x)$ and $f_1(x)$ enjoy the property that $f_1(x) > 0 \implies f_0(x) > 0$, (that is, the likelihood ratio does not "blow up" at any $x$), we have that $$E[\Lambda(X)\mid H_0] = \int_{\mathbb R}\frac{f_1(x)}{f_0(x)}\cdot f_0(x)\, \mathrm dx = \int_{\mathbb R}f_1(x)\, \mathrm dx = 1.$$

$$E[\Lambda^n(X)\mid H_1] = \int_{\mathbb R}\left[\frac{f_1(x)}{f_0(x)}\right]^n\cdot f_1(x)\, \mathrm dx = \int_{\mathbb R}\left[\frac{f_1(x)}{f_0(x)}\right]^{n+1}\cdot f_0(x)\, \mathrm dx = E[\Lambda^{n+1}(X)\mid H_0].$$

Dilip Sarwate
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