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I'm struggling with the following:

"A certain offer is available for maximum 2 days. The sellsman claims, the offer will be over at a random date. You think, the probability is decreasing linear from start to end. In fact, the offer is deleted after 8 hours."

a) Compute the density functions.

I've got $f_1=\frac{1}{2}$ (for the sellsman claim) and $f_2=-\frac{1}{2}x\bf{+1}$ (decreasing)

b) Frame a statistical test for the situation, if you want to accuse the sellsman wrongly with 20% probability.

If thought about an hypothesis testing with

$H_0$= the sellsman is right

$H_1$=I am right

$M_0=\lbrace c \leq x \leq 2\rbrace$ (valid area)

$M_1=\lbrace 0 \leq x \leq c \rbrace$ (rejection area)

and $\alpha=0.2$.

Now I started with \begin{eqnarray*}P(T \in M_1) \leq 0.2 &\Leftrightarrow& P(T \leq c) \leq \alpha \\ &\Leftrightarrow& \int \limits_0^c \frac{1}{2} dx \leq 0.2 \\ &\Leftrightarrow& \left[ \frac{1}{2} x \right]_0^c \leq 0.2 \\ &\Leftrightarrow& \frac{1}{2}c \leq 0.2 \Leftrightarrow c=0.4 \end{eqnarray*}

Is this correct so far?

Sorry for the bad english.

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Maybe would be more appropriate for this question. – Juris Jan 21 '13 at 12:29
Actually it is also asked (but is as yet unanswered) on stats.SE – Dilip Sarwate Jan 21 '13 at 15:33
up vote 1 down vote accepted

I think there are some problems at your problem formulation. First, at least to me, it is unclear what is meant by a random rate. Probably it means a uniform distribution for the range $[0,2]$ with $1/2$. Let this distribution be the null hypothesis as you mentioned and let it be the hypothesis that the salesman is right $H_0$.

The alternative hypothesis $H_1$ should be the hypothesis that you are right against the null hypothesis that the salesman is right. According to the question, the density should be decreasing but how? let it be decreasing linearly. Then you will have a triangle on $[0,2]$ whose area should sum up to $1$. $f_2$ which you thought of doesnt satisfy this property.

Next, given two densities $f_0$ and $f_1$, accusing the sellsman wrongly with $20\%$ corresponds to finding a suitable region $R$ in $[0,2]$ such that the area of $R$ under $H_0$ is $20\%$. Namely there will be some interval from where you integrate $f_0$ which will sum up to $0.2$. However there will be only a unique interval $R$ which will at the same time maximize the probability of detection; that is the probability that $H_1$ is true when actually $H_1$ is true. For this problem, that can be determined only with a unique threshold or similarly with a unique interval. Therefore the integral will start from $0$ till $c$. at this point you are correct $\rightarrow c=0.4$

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Yes you're absolutly right, I forgot some details in my formulation. Sorry about that. I added some. – ulead86 Jan 21 '13 at 13:46

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