0
$\begingroup$

Let $D={D_k}_1^N$ denotes a set of i.i.d observations, and x denotes random variables, p() denotes probability, whether $p(D|x)$ equals to $p(D)$? Thank you for any help.

EDIT: To be more clear, I talk this problem in Bayesian inference context where we seek the posterior distribution $p(x|D)~p(x)p(D|x)$, so D is dependent on x.

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
1
$\begingroup$

You stated that D are i.i.d. so they are independent to each other.

The crucial question is whether D are all independent of x.

p(D|x) = p(D)

only if D are all independent of x

Improve this answer
$\endgroup$
0
$\begingroup$

The general rule is the following:

X and Y are independent if either $P(X|Y)=P(X)$ or $P(Y|X)=P(Y)$

So if both, D and x, are independent, then $P(D|x)=P(D)$. I hope this answers your question. However, if you meant that D and x are i.i.d., then $P(D|x)=P(D)$ is true.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I think your last sentence confuses the answer. It is true, but OP has not stated any kind of relationship between $D$ and $x$ (e.g. whether $x$ is a new sample from same population, and actual sample from $D$, some label dependent on $D$ etc). This is worth clarifying with the OP $\endgroup$
    – Neil Slater
    Commented May 25, 2017 at 15:07
  • $\begingroup$ I did that by adding that sentence at the end of my answer (in case the OP meant D and x are i.i.d.). I even wrote "I hope this answers your question." beforehand, clearly indicating that the answer to the question asked is not the last sentence but what is mentioned before. But it is true, the question should be asked in a more specific way. $\endgroup$
    – Kuma
    Commented May 25, 2017 at 15:27
  • $\begingroup$ @Kuma X and Y are independent if either $p(x|y)=p(x)$ or $p(y|x)=p(y)$ $\endgroup$
    – Lella
    Commented Jun 9, 2017 at 7:25
  • $\begingroup$ That is true, I added that to my answer. Thanks for the input, @L.V.Rao ! $\endgroup$
    – Kuma
    Commented Jun 9, 2017 at 10:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.