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 March 21st, 2017, 04:05 PM #11 Newbie   Joined: Mar 2017 From: lyon Posts: 18 Thanks: 0 I use the Bayes formula: 1) ŽĆ(╬╗/x)= f(x|╬╗)*ŽĆ(╬╗) / f(x) and i find ŽĆ(╬╗/x)=exp(-x) for x positive 2) but for the second question, i find the same distribution function, does it correct? I think something is wrong with my calcul. March 22nd, 2017, 03:02 PM #12 Newbie   Joined: Mar 2017 From: lyon Posts: 18 Thanks: 0 Thanks again for your help, for the next exercisE, we have Consider the density f(x|╬╗) = 1/╬╗*exp(ŌłÆx) for x Ōēź 0. 1) If the prior for ╬╗ Ōēź 1 is uninformative i.e. ŽĆ(╬╗) ŌłØ 1 what is the posterior ? Is it proper ? 2) If the prior for ╬╗ Ōēź 1 is ŽĆ(╬╗) ŌłØ ╬╗^(ŌłÆ2) what is the posterior ? Is it proper ? Does it admit an expectation ? Does it admit a variance ? March 22nd, 2017, 07:45 PM   #13
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 Originally Posted by canfly I use the Bayes formula: 1) ŽĆ(╬╗/x)= f(x|╬╗)*ŽĆ(╬╗) / f(x) and i find ŽĆ(╬╗/x)=exp(-x) for x positive 2) but for the second question, i find the same distribution function, does it correct? I think something is wrong with my calcul.
I don't believe you've done (1) correctly.

How did you find $f(x)$ and what did you get for it?

Also are you sure the conditional distribution isn't

$f(x|\lambda) = \dfrac{e^{-\lambda x}}{\lambda}$ ? March 23rd, 2017, 06:08 AM   #14
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 Originally Posted by romsek I don't believe you've done (1) correctly. How did you find $f(x)$ and what did you get for it? Also are you sure the conditional distribution isn't $f(x|\lambda) = \dfrac{e^{-\lambda x}}{\lambda}$ ?
To compute $f(x)$( marginal distribution), I use the integral of f(x|\lambda)*the prior distribution. So I integer from 0 to + infini, (1/lambda)*e^{-\lambda x} dx and i get 1/lambda. If i replace it in the baysian formula, i will get exp(-x) for x positive.
For your question, i will ask my supervisor, may be you are right, you seem to be a very high expert in this field it's will look like to the poisson distribution or the exponential distribution if we put the lambda. March 23rd, 2017, 11:50 AM   #15
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 Originally Posted by canfly To compute $f(x)$( marginal distribution), I use the integral of f(x|\lambda)*the prior distribution. So I integer from 0 to + infini, (1/lambda)*e^{-\lambda x} dx and i get 1/lambda. If i replace it in the baysian formula, i will get exp(-x) for x positive. For your question, i will ask my supervisor, may be you are right, you seem to be a very high expert in this field it's will look like to the poisson distribution or the exponential distribution if we put the lambda.
this is what I get..

First off we have to assume some prior on $\lambda$

The problem specifies it's a uniform distribution and that $\lambda\geq 1$

so we'll let $\pi(\lambda) = \dfrac{1}{\alpha}, \lambda \in [1,1+\alpha],~\alpha>0$

$f_X(x) = \displaystyle \int_1^{1+\alpha}~\dfrac 1 \alpha \dfrac{e^{-x}}{\lambda}~d \lambda = \dfrac{e^{-x}\ln(1+\alpha)}{\alpha}$

$f_{\lambda|X}(x) = \dfrac{\dfrac{e^{-x}}{\lambda}\dfrac{1}{\alpha}}{\dfrac{e^{-x}\ln(1+\alpha)}{\alpha}} = \dfrac{1}{\lambda \ln(1+\alpha)},~\alpha>0,~\lambda \in [1,1+\alpha]$

This is a proper distribution. It's non-negative for it's entire range of support and it integrates across that range of support to 1.

I'll have to take a look at (2)

btw, It does look like $\dfrac{e^{-x}}{\lambda}$ is the proper conditional pdf.

The problem results in oddball functions if the $\lambda$ is included in the exponent.

Last edited by romsek; March 23rd, 2017 at 12:04 PM. March 23rd, 2017, 12:03 PM   #16
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 Originally Posted by canfly Thanks again for your help, for the next exercisE, we have Consider the density f(x|╬╗) = 1/╬╗*exp(ŌłÆx) for x Ōēź 0. 2) If the prior for ╬╗ Ōēź 1 is ŽĆ(╬╗) ŌłØ ╬╗^(ŌłÆ2) what is the posterior ? Is it proper ? Does it admit an expectation ? Does it admit a variance ?
(2) Is pretty straightforward.

$\pi(\lambda) = \dfrac{1}{\lambda^2},~\lambda \in [1,\infty)$

a straightforward integration gets you

$f_X(x) = \dfrac {e^{-x}}{2},~0\leq x$

This results in

$f_{\lambda|X} = \dfrac{2}{\lambda^3},~\lambda \in [1,\infty)$

$E[\lambda|X] = 2$

The distribution does not admit a variance as it's 2nd moment diverges. March 23rd, 2017, 01:46 PM   #17
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 Originally Posted by romsek (2) Is pretty straightforward. $\pi(\lambda) = \dfrac{1}{\lambda^2},~\lambda \in [1,\infty)$ a straightforward integration gets you $f_X(x) = \dfrac {e^{-x}}{2},~0\leq x$ This results in $f_{\lambda|X} = \dfrac{2}{\lambda^3},~\lambda \in [1,\infty)$ $E[\lambda|X] = 2$ The distribution does not admit a variance as it's 2nd moment diverges.
Thanks again, but I don't understand why did you take for the prior a uniforme distribution by choising a alpha? in the exercise, they say it's egal to 1 or i miss understand something.
For the 2), you take 1/lambda^2 , here why don't you take a alpha like in the question 1), it's the deux points that i don't unsdertand after for te calcul i understand. March 23rd, 2017, 02:58 PM   #18
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 Originally Posted by canfly Thanks again, but I don't understand why did you take for the prior a uniforme distribution by choising a alpha? in the exercise, they say it's egal to 1 or i miss understand something. For the 2), you take 1/lambda^2 , here why don't you take a alpha like in the question 1), it's the deux points that i don't unsdertand after for te calcul i understand.
the $\propto$ symbol means proportional to not equal to.

In this case it just means that the PDF is constant i.e. a uniform distribution.

Well, it's got to be uniform over some interval.

In the second case if the distribution is $\dfrac {c}{\lambda^2}$ the fact that

$\displaystyle \int_1^\infty~\dfrac {c}{\lambda^2}~d\lambda = 1 \Rightarrow c = 1$ March 24th, 2017, 02:44 AM   #19
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 Originally Posted by romsek the $\propto$ symbol means proportional to not equal to. In this case it just means that the PDF is constant i.e. a uniform distribution. Well, it's got to be uniform over some interval. In the second case if the distribution is $\dfrac {c}{\lambda^2}$ the fact that $\displaystyle \int_1^\infty~\dfrac {c}{\lambda^2}~d\lambda = 1 \Rightarrow c = 1$
Yes,very clear ;thans you again, are you a teacher in this field? For the lamba in the exponential, i am wainting the answer of my supervisor. I think, it's not change the step of your reasoning, is it? March 24th, 2017, 04:53 AM   #20
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 Originally Posted by canfly For the lamba in the exponential, i am wainting the answer of my supervisor. I think, it's not change the step of your reasoning, is it?
The problem is correct as written.

I'm just a retired engineer who actually used some of this stuff from time to time. Tags baysian, econometrics Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Ku5htr1m Economics 0 March 11th, 2016 11:36 AM sarwar Probability and Statistics 0 September 19th, 2014 09:27 AM BartL Math Books 0 October 25th, 2010 11:17 PM princessanna57 Economics 1 September 11th, 2009 06:29 PM

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