March 21st, 2017, 04:05 PM  #11 
Newbie Joined: Mar 2017 From: lyon Posts: 18 Thanks: 0 
I use the Bayes formula: 1) π(λ/x)= f(xλ)*π(λ) / f(x) and i find π(λ/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(xλ) = 1/λ*exp(−x) for x ≥ 0. 1) If the prior for λ ≥ 1 is uninformative i.e. π(λ) ∝ 1 what is the posterior ? Is it proper ? 2) If the prior for λ ≥ 1 is π(λ) ∝ λ^(−2) what is the posterior ? Is it proper ? Does it admit an expectation ? Does it admit a variance ? 
March 22nd, 2017, 07:45 PM  #13  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,484 Thanks: 744  Quote:
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  
Newbie Joined: Mar 2017 From: lyon Posts: 18 Thanks: 0  Quote:
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  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,484 Thanks: 744  Quote:
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 $\pi(\lambda) = \dfrac{1}{\alpha}, \lambda \in [1,1+\alpha],~\alpha>0$ $f_X(x) = \displaystyle \int_1^{1+\alpha}~\dfrac 1 \alpha \dfrac{e^{x}}{\lambda}~d \lambda = \dfrac{e^{x}\ln(1+\alpha)}{\alpha}$ $f_{\lambdaX}(x) = \dfrac{\dfrac{e^{x}}{\lambda}\dfrac{1}{\alpha}}{\dfrac{e^{x}\ln(1+\alpha)}{\alpha}} = \dfrac{1}{\lambda \ln(1+\alpha)},~\alpha>0,~\lambda \in [1,1+\alpha]$ This is a proper distribution. It's nonnegative 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  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,484 Thanks: 744  Quote:
$\pi(\lambda) = \dfrac{1}{\lambda^2},~\lambda \in [1,\infty)$ a straightforward integration gets you $f_X(x) = \dfrac {e^{x}}{2},~0\leq x$ This results in $f_{\lambdaX} = \dfrac{2}{\lambda^3},~\lambda \in [1,\infty)$ $E[\lambdaX] = 2$ The distribution does not admit a variance as it's 2nd moment diverges.  
March 23rd, 2017, 01:46 PM  #17  
Newbie Joined: Mar 2017 From: lyon Posts: 18 Thanks: 0  Quote:
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  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,484 Thanks: 744  Quote:
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  
Newbie Joined: Mar 2017 From: lyon Posts: 18 Thanks: 0  Quote:
 
March 24th, 2017, 04:53 AM  #20 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,484 Thanks: 744  

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baysian, econometrics 
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