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Sorry for the really simple question. I am reading a book about psychophysiology and I got lost at one point. I will take out all the unrelevant information and just go straight to the point. The book present this probabilities:

1) f(r,t|E)=Pr(r=r, T=t| E); where E, from now on, is defined as a vector of variables.

2) g(x|E)=Pr(X=x|E);

3) h(r,t|x)=Pr(r=r, T=t| x);

and finally it shows that "by the law of total probability applied to the conditional probability"

4) $$f(r,t|E)=\int g(x|E)*h(r,t|x) dx.$$

How can I derive 4 from 1-2-3?

In the text it is said that (r,t) has no direct connection with E exepct as it is mediated by the parameter X. I suspect that this means that (r,t) and E are independent, but (r,t) and X are not. I also guess that X and E dependent. Is that correct?

Vaaal88
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1 Answers1

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It is not saying $(R,T)$ and $E$ are independent, but that they are conditionally independent given $X=x$. So it is saying $\Pr(R=r, T=t|X=x,E=e)=\Pr(R=r, T=t|X=x)$.

You are then integrating over all $x$ to remove the nuisance parameter.

You have the conditional probability statement $\Pr(A=a,B=b)=\Pr(A=a|B=b)\Pr(B=b)$ which you can extend to $\Pr(A=a,B=b|C=c)=\Pr(A=a|B=b,C=c)\Pr(B=b|C=c)$. Then summing over $b$ you get $$\Pr(A=a|C=c) = \sum_b \Pr(A=a,B=b|C=c)$$ $$=\sum_b \Pr(A=a|B=b,C=c)\Pr(B=b|C=c).$$

Now you need to replace $A=a$ by $R=r, T=t$; replace $B=b$ by $X=x$; and replace $C=c$ by $E=e$. Then use the conditional independence statement above. Finally move from a sum of conditional probabilities to an integral of conditional probability densities with functions $f,g,h$.

Henry
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