Bayesian probability: proving $P(x|y,x) = 1$












0












$begingroup$


I have an exercise where I'm supposed to prove that $P(x|y,x) = 1$



I've come up with the following but I'm not sure if it feels right to me:



$P(x|y,x) =$



$P(x,y,x) / P(y,x) = $



$P(x|y,x)P(y|x)P(x) / P(y,x) = $



$P(x|y,x)P(y|x)P(x) / P(y|x)P(x) = $



We now have



$P(x|y,x) = P(x|y,x)$



So:



$P(x|y,x) / P(x|y,x) = P(x|y,x) / P(x|y,x) = 1$



Is this correct?






bayesian bayes-theorem







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  • $begingroup$
    You question is not clear. What does $p(x,y,x)$ mean?
    $endgroup$
    – user144410
    Jan 16 at 13:27












  • $begingroup$
    Sorry, it might be an unusual notation used in my class, it means P(x∩y∩x)
    $endgroup$
    – ronaldfisher
    Jan 16 at 13:40


















0












$begingroup$


I have an exercise where I'm supposed to prove that $P(x|y,x) = 1$



I've come up with the following but I'm not sure if it feels right to me:



$P(x|y,x) =$



$P(x,y,x) / P(y,x) = $



$P(x|y,x)P(y|x)P(x) / P(y,x) = $



$P(x|y,x)P(y|x)P(x) / P(y|x)P(x) = $



We now have



$P(x|y,x) = P(x|y,x)$



So:



$P(x|y,x) / P(x|y,x) = P(x|y,x) / P(x|y,x) = 1$



Is this correct?






bayesian bayes-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    You question is not clear. What does $p(x,y,x)$ mean?
    $endgroup$
    – user144410
    Jan 16 at 13:27












  • $begingroup$
    Sorry, it might be an unusual notation used in my class, it means P(x∩y∩x)
    $endgroup$
    – ronaldfisher
    Jan 16 at 13:40
















0












0








0





$begingroup$


I have an exercise where I'm supposed to prove that $P(x|y,x) = 1$



I've come up with the following but I'm not sure if it feels right to me:



$P(x|y,x) =$



$P(x,y,x) / P(y,x) = $



$P(x|y,x)P(y|x)P(x) / P(y,x) = $



$P(x|y,x)P(y|x)P(x) / P(y|x)P(x) = $



We now have



$P(x|y,x) = P(x|y,x)$



So:



$P(x|y,x) / P(x|y,x) = P(x|y,x) / P(x|y,x) = 1$



Is this correct?






bayesian bayes-theorem







share|cite|improve this question











$endgroup$




I have an exercise where I'm supposed to prove that $P(x|y,x) = 1$



I've come up with the following but I'm not sure if it feels right to me:



$P(x|y,x) =$



$P(x,y,x) / P(y,x) = $



$P(x|y,x)P(y|x)P(x) / P(y,x) = $



$P(x|y,x)P(y|x)P(x) / P(y|x)P(x) = $



We now have



$P(x|y,x) = P(x|y,x)$



So:



$P(x|y,x) / P(x|y,x) = P(x|y,x) / P(x|y,x) = 1$



Is this correct?






bayesian bayes-theorem




bayesian bayes-theorem






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share|cite|improve this question













share|cite|improve this question




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edited Jan 16 at 13:31







ronaldfisher

















asked Jan 16 at 13:21









ronaldfisherronaldfisher

11




11












  • $begingroup$
    You question is not clear. What does $p(x,y,x)$ mean?
    $endgroup$
    – user144410
    Jan 16 at 13:27












  • $begingroup$
    Sorry, it might be an unusual notation used in my class, it means P(x∩y∩x)
    $endgroup$
    – ronaldfisher
    Jan 16 at 13:40




















  • $begingroup$
    You question is not clear. What does $p(x,y,x)$ mean?
    $endgroup$
    – user144410
    Jan 16 at 13:27












  • $begingroup$
    Sorry, it might be an unusual notation used in my class, it means P(x∩y∩x)
    $endgroup$
    – ronaldfisher
    Jan 16 at 13:40


















$begingroup$
You question is not clear. What does $p(x,y,x)$ mean?
$endgroup$
– user144410
Jan 16 at 13:27






$begingroup$
You question is not clear. What does $p(x,y,x)$ mean?
$endgroup$
– user144410
Jan 16 at 13:27














$begingroup$
Sorry, it might be an unusual notation used in my class, it means P(x∩y∩x)
$endgroup$
– ronaldfisher
Jan 16 at 13:40






$begingroup$
Sorry, it might be an unusual notation used in my class, it means P(x∩y∩x)
$endgroup$
– ronaldfisher
Jan 16 at 13:40












1 Answer
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$begingroup$

Assuming that $P(x|y,x) = P(x|y cap x)$, just note that
$$x cap y cap x = y cap x Rightarrow P(x|y cap x) = frac{P(x cap y cap x)}{P(y cap x)} = frac{P(y cap x)}{P(y cap x)} = 1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You assumed right. Thank you
    $endgroup$
    – ronaldfisher
    Jan 16 at 13:36











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Assuming that $P(x|y,x) = P(x|y cap x)$, just note that
$$x cap y cap x = y cap x Rightarrow P(x|y cap x) = frac{P(x cap y cap x)}{P(y cap x)} = frac{P(y cap x)}{P(y cap x)} = 1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You assumed right. Thank you
    $endgroup$
    – ronaldfisher
    Jan 16 at 13:36
















0












$begingroup$

Assuming that $P(x|y,x) = P(x|y cap x)$, just note that
$$x cap y cap x = y cap x Rightarrow P(x|y cap x) = frac{P(x cap y cap x)}{P(y cap x)} = frac{P(y cap x)}{P(y cap x)} = 1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You assumed right. Thank you
    $endgroup$
    – ronaldfisher
    Jan 16 at 13:36














0












0








0





$begingroup$

Assuming that $P(x|y,x) = P(x|y cap x)$, just note that
$$x cap y cap x = y cap x Rightarrow P(x|y cap x) = frac{P(x cap y cap x)}{P(y cap x)} = frac{P(y cap x)}{P(y cap x)} = 1$$






share|cite|improve this answer









$endgroup$



Assuming that $P(x|y,x) = P(x|y cap x)$, just note that
$$x cap y cap x = y cap x Rightarrow P(x|y cap x) = frac{P(x cap y cap x)}{P(y cap x)} = frac{P(y cap x)}{P(y cap x)} = 1$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 16 at 13:32









trancelocationtrancelocation

12.1k1826




12.1k1826












  • $begingroup$
    You assumed right. Thank you
    $endgroup$
    – ronaldfisher
    Jan 16 at 13:36


















  • $begingroup$
    You assumed right. Thank you
    $endgroup$
    – ronaldfisher
    Jan 16 at 13:36
















$begingroup$
You assumed right. Thank you
$endgroup$
– ronaldfisher
Jan 16 at 13:36




$begingroup$
You assumed right. Thank you
$endgroup$
– ronaldfisher
Jan 16 at 13:36


















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