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If one coin toss yields Head, what is the PDF for the probability of a head?

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2

$begingroup$

This is not a homework question. I discovered that I am not the only tenant in a highrise that is getting headaches from paint fumes. I talked to one couple in the elevator about the paint fumes. I turns out that they have the same problem. Based on just one observation, I would be dismissed as a quack if I try to convince the property manager that tenants in general are likely to find the paint fumes problematic. However, I was wondering if that one observation allowed one to contruct a PDF of $p$ with a broad spread ($p$ being the proportion of tenants who are troubled by the paint fumes).

probability

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asked 2 days ago

user2153235user2153235

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  If you get an inspector in to measure the level of toxic fumes, you do not need statistics.
  $endgroup$
  – Carl
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I wasn't going to present the statistics, it was an intellectual question. The city is in-bed with the large owners of many highrises in the city, and one must be prepared to quit one's job and push full time if due diligence is desired. I'm currently appealing to the property owners desire for attractive premises, and the hurdle right now is to convince them that the observed periods of no ventilation are reliable. Hopefully, a few witnesses will do that.
  $endgroup$
  – user2153235
  5 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

This is not a homework question. I discovered that I am not the only tenant in a highrise that is getting headaches from paint fumes. I talked to one couple in the elevator about the paint fumes. I turns out that they have the same problem. Based on just one observation, I would be dismissed as a quack if I try to convince the property manager that tenants in general are likely to find the paint fumes problematic. However, I was wondering if that one observation allowed one to contruct a PDF of $p$ with a broad spread ($p$ being the proportion of tenants who are troubled by the paint fumes).

probability

share|cite|improve this question

asked 2 days ago

user2153235user2153235

132

New contributor

user2153235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  If you get an inspector in to measure the level of toxic fumes, you do not need statistics.
  $endgroup$
  – Carl
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I wasn't going to present the statistics, it was an intellectual question. The city is in-bed with the large owners of many highrises in the city, and one must be prepared to quit one's job and push full time if due diligence is desired. I'm currently appealing to the property owners desire for attractive premises, and the hurdle right now is to convince them that the observed periods of no ventilation are reliable. Hopefully, a few witnesses will do that.
  $endgroup$
  – user2153235
  5 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This is not a homework question. I discovered that I am not the only tenant in a highrise that is getting headaches from paint fumes. I talked to one couple in the elevator about the paint fumes. I turns out that they have the same problem. Based on just one observation, I would be dismissed as a quack if I try to convince the property manager that tenants in general are likely to find the paint fumes problematic. However, I was wondering if that one observation allowed one to contruct a PDF of $p$ with a broad spread ($p$ being the proportion of tenants who are troubled by the paint fumes).

probability

share|cite|improve this question

asked 2 days ago

user2153235user2153235

132

New contributor

user2153235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked 2 days ago

user2153235user2153235

132

New contributor

user2153235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 2 days ago

user2153235user2153235

132

New contributor

user2153235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 days ago

user2153235user2153235

132

New contributor

user2153235 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 days ago

user2153235user2153235

132

1 Answer
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2

$begingroup$

Independent of paint fumes, the question can be thought of figuring out if a coin is biased or not, based on one (or more if you prefer) toss. One logical way to make inference is using Bayesian framework, but we need priors. Just as in example in here, assuming a non-informative prior, i.e. uniform distribution, which is the same with beta distribution with $alpha=1,beta=1$, the outcome of the experiment can be incorporated into the PDF estimation, yo come up with a posterior. The posterior is again Beta distributed, with $alpha=alpha_0+s$, $beta=beta_0+f$, where $alpha_0,beta_0$ represent your initial belief, i.e. prior, and $s$ number of positive outcomes, and $f$ number of negative outcomes after $n$ trials. In your case, $n=1,f=0,s=1$. And, when you select the non-informative prior, i.e. $alpha_0=1,beta_0=1$, you end up with a Beta distribution for $p$ with $alpha=2,beta=1$:

share|cite|improve this answer

answered 2 days ago

gunesgunes

7,8011317

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks gunes. I was hoping that it would be a canned equation instead of Bayesian analysis, but you made it both. I tried to crack the Bayesian nut multiple times in the past half decade, but never got my head around it. Reading things like Bayesian inference, Think Bayes, and How to measure anything (Chapt.10). Your answer inspired me to try the last one again. Together with your example, I am getting a foothold.
  $endgroup$
  – user2153235
  4 hours ago
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

Independent of paint fumes, the question can be thought of figuring out if a coin is biased or not, based on one (or more if you prefer) toss. One logical way to make inference is using Bayesian framework, but we need priors. Just as in example in here, assuming a non-informative prior, i.e. uniform distribution, which is the same with beta distribution with $alpha=1,beta=1$, the outcome of the experiment can be incorporated into the PDF estimation, yo come up with a posterior. The posterior is again Beta distributed, with $alpha=alpha_0+s$, $beta=beta_0+f$, where $alpha_0,beta_0$ represent your initial belief, i.e. prior, and $s$ number of positive outcomes, and $f$ number of negative outcomes after $n$ trials. In your case, $n=1,f=0,s=1$. And, when you select the non-informative prior, i.e. $alpha_0=1,beta_0=1$, you end up with a Beta distribution for $p$ with $alpha=2,beta=1$:

share|cite|improve this answer

answered 2 days ago

gunesgunes

7,8011317

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks gunes. I was hoping that it would be a canned equation instead of Bayesian analysis, but you made it both. I tried to crack the Bayesian nut multiple times in the past half decade, but never got my head around it. Reading things like Bayesian inference, Think Bayes, and How to measure anything (Chapt.10). Your answer inspired me to try the last one again. Together with your example, I am getting a foothold.
  $endgroup$
  – user2153235
  4 hours ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Independent of paint fumes, the question can be thought of figuring out if a coin is biased or not, based on one (or more if you prefer) toss. One logical way to make inference is using Bayesian framework, but we need priors. Just as in example in here, assuming a non-informative prior, i.e. uniform distribution, which is the same with beta distribution with $alpha=1,beta=1$, the outcome of the experiment can be incorporated into the PDF estimation, yo come up with a posterior. The posterior is again Beta distributed, with $alpha=alpha_0+s$, $beta=beta_0+f$, where $alpha_0,beta_0$ represent your initial belief, i.e. prior, and $s$ number of positive outcomes, and $f$ number of negative outcomes after $n$ trials. In your case, $n=1,f=0,s=1$. And, when you select the non-informative prior, i.e. $alpha_0=1,beta_0=1$, you end up with a Beta distribution for $p$ with $alpha=2,beta=1$:

share|cite|improve this answer

answered 2 days ago

gunesgunes

7,8011317

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks gunes. I was hoping that it would be a canned equation instead of Bayesian analysis, but you made it both. I tried to crack the Bayesian nut multiple times in the past half decade, but never got my head around it. Reading things like Bayesian inference, Think Bayes, and How to measure anything (Chapt.10). Your answer inspired me to try the last one again. Together with your example, I am getting a foothold.
  $endgroup$
  – user2153235
  4 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Independent of paint fumes, the question can be thought of figuring out if a coin is biased or not, based on one (or more if you prefer) toss. One logical way to make inference is using Bayesian framework, but we need priors. Just as in example in here, assuming a non-informative prior, i.e. uniform distribution, which is the same with beta distribution with $alpha=1,beta=1$, the outcome of the experiment can be incorporated into the PDF estimation, yo come up with a posterior. The posterior is again Beta distributed, with $alpha=alpha_0+s$, $beta=beta_0+f$, where $alpha_0,beta_0$ represent your initial belief, i.e. prior, and $s$ number of positive outcomes, and $f$ number of negative outcomes after $n$ trials. In your case, $n=1,f=0,s=1$. And, when you select the non-informative prior, i.e. $alpha_0=1,beta_0=1$, you end up with a Beta distribution for $p$ with $alpha=2,beta=1$:

share|cite|improve this answer

answered 2 days ago

gunesgunes

7,8011317

$endgroup$

share|cite|improve this answer

answered 2 days ago

gunesgunes

7,8011317

answered 2 days ago

gunesgunes

7,8011317

answered 2 days ago

gunesgunes

7,8011317