Hint: Do you have the relationship between correlation and the joint density functions?
I have just started learning these things, would really appreciate any help on this.
There are 4 nodes in a network (say A,B,C,D)and each node has certain failure probabilities( P(af),P(bf),P(cf),P(df) ).
Then,I have this 4x4 correlation Matrix which contains the correlation coefficients for every pair of nodes.
How can I calculate P(af|bf cf df)?
Thanks for the reply. I am not sure, but I think the values in the joint density function could be considered the same as the values from the correlation Matrix for every pair of nodes, but I have difficulties doing it for more than two random variables. Could anyone please reply?
Thanks for the reply again.
I am sorry if this is wrong, but this is what I think is OK to do, at least the application in which I am using it.
Corr(a,b) = P(a|b) = P(b|a). I think there would be no problem to use this in my case. Now, a also has correlation with c & d which is again given by Corr(a,c) and Corr(a,d). So, I need to find P(a|b,c,d), from all these values. Again, I am very new at this, and please bear with me if it seems incorrect.
In general, P(a|b) != P(b|a): they are only equal if P(A) = P(B).
Is this from a class or this something else? If this is from a class, then what kind of identities/formulae do you have?
There are expressions that relate correlations to probabilities but I think asking you the above question will move us towards rectifying the problem.
Thanks for the reply again. Well, I know in general P(a|b) != P(b|a). I am a little confused though.
This is a small part of what I am trying to do for my class presentation. Maybe, I will just clearly explain the whole problem. I really appreciate your replies to this.
So, there are 4 nodes in the network and each node has certain failure probability which we know and are different. Now, the correlation, which is determined by the distance between the two nodes, should be such that if one node fails in the network, the other node fails depending on the amount of corrrelation(depedning on the distance again) . i.e. if node 1 fails,then there is a higher probability of node 2 failing as node 2 is closer to node 1 than node 3 and node 4. Please see the attached file probability.png