Prediction with cpdist using "probabilities" as evidence

Question

I have a very quick question with an easy reproducible example that is related to my work on prediction with bnlearn

    library(bnlearn)
    Learning.set4=cbind(c("Yes","Yes","Yes","No","No","No"),c(9,10,8,3,2,1))
    Learning.set4=as.data.frame(Learning.set4)
    Learning.set4[,c(2)]=as.numeric(as.character(Learning.set4[,c(2)]))
    colnames(Learning.set4)=c("Cause","Cons")
    b.network=empty.graph(colnames(Learning.set4))
    struct.mat=matrix(0,2,2)
    colnames(struct.mat)=colnames(Learning.set4)
    rownames(struct.mat)=colnames(struct.mat)
    struct.mat[1,2]=1
    bnlearn::amat(b.network)=struct.mat
    haha=bn.fit(b.network,Learning.set4)


    #Some predictions with "lw" method

    #Here is the approach I know with a SET particular modality. 
    #(So it's happening with certainty, here for example I know Cause is "Yes")
    classic_prediction=cpdist(haha,nodes="Cons",evidence=list("Cause"="Yes"),method="lw")
    print(mean(classic_prediction[,c(1)]))


    #What if I wanted to predict the value of Cons, when Cause has a 60% chance of being Yes and 40% of being no?
    #I decided to do this, according the help
    #I could also make a function that generates "Yes" or "No" with proper probabilities.
    prediction_idea=cpdist(haha,nodes="Cons",evidence=list("Cause"=c("Yes","Yes","Yes","No","No")),method="lw")
    print(mean(prediction_idea[,c(1)]))

Here is what the help says:

"In the case of a discrete or ordinal node, two or more values can also be provided. In that case, the value for that node will be sampled with uniform probability from the set of specified values"

When I predict the value of a variable using categorical variables, I for now just used a certain modality of said variable as in the first prediction in the example. (Having the evidence set at "Yes" gets Cons to take a high value)

But if I wanted to predict Cons without knowing the exact modality of the variable Cause with certainty, could I use what I did in the second prediction (Just knowing the probabilities) ? Is this an elegant way or are there better implemented ones I don't know off?

Joel ; not related to your question but maybe you can help: from your example, `cpdist(haha,nodes="Cons",evidence=list("Cause"="No"),method="lw")` this would seem to give (the expected by me) mean of two, from looking at the cpd's of your example, but why when `"Cause"="Yes")` is the mean ~zero? (i would of expected nine) — user20650, Jan 03 '17 at 19:52
Hello user20650. Can you post the full code that makes you get the mean of 0? By defining the network as above and then doing this `mean(cpdist(haha,nodes="Cons",evidence=list("Cause"="Yes"),method="lw")[,c(1)])` I get the expected mean around 9 — Joel H, Jan 04 '17 at 08:23

Joel H · Answer 1 · 2017-01-04T09:09:35.367

I got in touch with the creator of the package, and I will paste his answer related to the question here:

The call to cpquery() is wrong:

Prediction_idea=cpdist(haha,nodes="Cons",evidence=list("Cause"=c("Yes","Yes","Yes","No","No")),method="lw")
print(mean(prediction_idea[,c(1)]))

A query with the 40%-60% soft evidence requires you to place these new probabilities in the network first

haha$Cause = c(0.40, 0.60)

and then run the query without an evidence argument. (Because you do not have any hard evidence, really, just a different probability distribution for Cause.)

I will post the code that lets me do what I wanted off of the fitted network from the example.

change=haha$Cause$prob
change[1]=0.4
change[2]=0.6
haha$Cause=change
new_prediction=cpdist(haha,nodes="Cons",evidence=TRUE,method="lw")
print(mean(new_prediction[,c(1)]))

Prediction with cpdist using "probabilities" as evidence

1 Answers1