2 fixed some notation

Suppose you perform 2 draws and observe the sequence {v1, 0}. Then your question is:

P( {v1, 0} | m ) = ?.

The intuitive way to think of P( {v1, 0} | m ) is the take the frequentist perspective and imagine that we perform 2 draws N times where N is a very large number and count the number of times we observe {v1, 0}. Then, as N tends to infinity, we have

P( {v1, 0} | m ) = No of times we observe {v1, 0} / N

If you imagine doing this experiment then "No of times we observe {v1, 0} / N" will clearly equal f(v1 | v1 ~ N(m,1) I(v1 >0) m) * Prob( v1 <0 v < 0 | m ) where v1 ~ N(m,1) I(v1 >0) ) and where v ~ N(m,1).

To see why the above is correct, note that: (a) the two events {v1} and {0} are independent, (b) the number of times you draw v1 is proportional to v1 being drawn from a truncated normal and (c) the number of times you see 0 is proportional to the probability that a normal draw is a negative number.

Estimating m is then a matter of using either maximum-likelihood or bayesian ideas.

Hope that helps.

1

Suppose you perform 2 draws and observe the sequence {v1, 0}. Then your question is:

P( {v1, 0} | m ) = ?.

The intuitive way to think of P( {v1, 0} | m ) is the take the frequentist perspective and imagine that we perform 2 draws N times where N is a very large number and count the number of times we observe {v1, 0}. Then, as N tends to infinity, we have

P( {v1, 0} | m ) = No of times we observe {v1, 0} / N

If you imagine doing this experiment then "No of times we observe {v1, 0} / N" will clearly equal f(v1 | v1 ~ N(m,1) I(v1 >0) ) Prob( v1 <0 | v1 ~ N(m,1) ).

To see why the above is correct, note that: (a) the two events {v1} and {0} are independent, (b) the number of times you draw v1 is proportional to v1 being drawn from a truncated normal and (c) the number of times you see 0 is proportional to the probability that a normal draw is a negative number.

Estimating m is then a matter of using either maximum-likelihood or bayesian ideas.

Hope that helps.