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I'm new to this topic.

Suppose that

  • $ f(x) = x^Tw$, where both $x$ and $w$ are independent random variables with known probability density function.

  • $ y(x) = f(x) + \epsilon $ where $ \epsilon \sim N(0, \sigma^2_n)$, and $\epsilon$ is independent of $x$ and $w$.

I would like to show that $$ p(w|y,x) = \frac{p(y|x,w) p(w)}{p(y|x)}. $$

Here is what I have tried:

problem equation

user144410
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Trainer99
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    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Aug 24 '18 at 22:01
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    You'll find that simple "Here's the statement of my exercise, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Aug 24 '18 at 22:01
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    Here's a MathJax tutorial :) – Shaun Aug 24 '18 at 22:02
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    Thanks !! @shaun – Trainer99 Aug 24 '18 at 22:09
  • I dont see the question. what do you want to prove – user144410 Aug 24 '18 at 22:13
  • Hi @user144410 can't you see the attached picture? – Trainer99 Aug 24 '18 at 22:22
  • I can see it, but there is not enough information. What are the assumptions on x, y, and w. Is w independent of x for example? – user144410 Aug 24 '18 at 22:34
  • maybe yes... w is independent of x and $ w \sim N(0,\sum_p)$ – Trainer99 Aug 24 '18 at 23:08
  • maybe yes... w is independent of x and $ w \sim N(0,\sum_p)$ Like the picture, i try to prove the equation using bayesian equation. However I failed. – Trainer99 Aug 24 '18 at 23:15

1 Answers1

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We have $$p(x,y,w) = p(w | x ,y) p(x,y) = p(w | x ,y) p(y|x) p(x) $$ and because $w$ and $x$ are independent $p(w|x) = p(w)$, and it holds that $$ p(x,y,w) = p(y | x ,w) p(w|x) p(x) = p(y | x ,w) p(w) p(x) $$

Therefore, equating the right hand sides, you get the desired expression.

user144410
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  • Thank you very much.. The important thing is whether w is independent of x or not. So, I have searched the relation between x and w. please check the main contents. – Trainer99 Aug 25 '18 at 02:05
  • Hi @user144410, I completely understand it.!!!

    You're right. $x$ is independent of $w$ even though $w$ and $x$ are dependent on $y$ because of $y = x^Tw + \epsilon$

     – Trainer99 Aug 25 '18 at 02:15
  • @Trainer99 You are welcome ! – user144410 Aug 26 '18 at 17:26