I recently went through a video which said that in the relation x->W<-Y, X does not influence y.X has causal relationship to W and W has evidential relationship to Y .So will X not affect Y ?
Why do v structures not contribute to flow of probabilistic influence?
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1[(Hey) Where did $v$ go?](http://vimeo.com/10738004) Do mean $W\leftarrow Y$ or $W<-Y$? I would definitely not hurt to provide some detail (including a link to the video if possible)... and welcome to MathStackExchange. – 2012-04-23
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0hey this is the video link : http://www.youtube.com/watch?v=PfirsYouObw – 2012-04-23
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0and I actually wanted to say X (arrow mark)-> W←Y – 2012-04-23
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0In your post, you interchange small and capital letters. For a reason? – 2012-04-23
1 Answers
Call $p(x)$ the probability to observe $X=x$, $q(y)$ the probability to observe $Y=y$, and $r(w\mid x,y)$ the probability to observe $W=w$ conditionally on $[X=x,Y=y]$. Assume that $X$ and $Y$ are independent.
The probability that $[X=x,Y=y]$ is $p(x)q(y)$ and the probability that $[X=x,Y=y,W=w]$ is $$ P(x,y,w)=p(x)q(y)r(w\mid x,y). $$ The probability that $[X=x,Y=y]$ conditionally on $[W=w]$ is $$ P(x,y\mid w)=\frac{P(x,y,w)}{\sum\limits_{x',y'}P(x',y',w)}=p(x)q(y)R(w,x,y), $$ where $$ R(w,x,y)=\frac{r(w\mid x,y)}{\sum\limits_{x',y'}p(x')q(y')r(w\mid x',y')}. $$ Since $R(w,x,y)$ is not, in general, a function of $(x,w)$ times a function of $(y,w)$, one sees that $X$ and $Y$ are not independent conditionally on $W$.
On the other hand, the probability that $X=x$ conditionally on $[Y=y]$ is $$ \frac{\sum\limits_wP(x,y,w)}{\sum\limits_{x',w}P(x',y,w)}=\frac{p(x)q(y)\sum\limits_wr(w\mid x,y)}{q(y)\sum\limits_{x'}p(x')\sum\limits_wr(w\mid x',y)}=p(x), $$ hence $X$ and $Y$ are indeed independent (this was a hypothesis of the model).