Re: Expected Value

On Jun 8, 5:40 am, "David Jones" <dajx...@xxxxxxxxx> wrote:
jay wrote:
On Jun 7, 11:20 am, DanCT <danmye...@xxxxxxxxxxx> wrote:
On Jun 6, 11:15 am, jay <jayesh.srivast...@xxxxxxxxx> wrote:

On Jun 6, 10:51 am, DanCT <danmye...@xxxxxxxxxxx> wrote:

I would like to compute the expectation of a continuous quantity X
where:
X = X(x) if T = 1, for T a binary value
X = 0 otherwise

X(x) will be fit using regression. The tricky part here is that T
is not known when scoring a population, so I am thinking I'll need
to resort to predicting a probability for T(t) = 1 using a logit
model.

So how do I compute E(X) in this case?

E(X) = p*E(X) where p is Pr(T = 1) which basically you are trying to
estimate from your modeling.

Isn't that only true if X & T are independent? (which they are not in
my case)

In your case X = 0 otherwise.

E(X) = Pr(T=1)*E(X) + Pr(T=0)*0 = p*E(X)

It would be better to use some proper notation throughout. The above is better written as the following, which
shows the role of dependence:

E(X) = Pr(T=1)*E(X|T=1) + Pr(T=0)*E(X|T=0) = p*E(X|T=1).

David Jones- Hide quoted text -

- Show quoted text -

Here's the thing though. If you right down the expression for X for
any given observation, it looks like:

X = I(T=1)* X|T=1 , where I(.) is an indicator

Taking the expectation gives:

E[X] = E[I(T=1)* X|T=1]

But,

E[I(T=1)* X|T=1] = E[I(T=1)] * E[X|T=1] {= P(T=1)*E[X|T=1], your
expression} only if X & T are independent.

See my point?


.

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