f(x) = 8/(x^3) x>2
the question seems simple enough
find E[X given X>3]
but with conditional probability it would be:
E(X intersect X>3)/E(X>3)
but X intersect X>3 would just be X>3
so it becomes E(X>3)/E(X>3) which would just equal 1
I think I must be going about this problem all wrong but I don't know how to solve this, any help?
O wait, would it just be 1 minus the probability of being between 2 and 3? Or is that not right either?
the question seems simple enough
find E[X given X>3]
but with conditional probability it would be:
E(X intersect X>3)/E(X>3)
but X intersect X>3 would just be X>3
so it becomes E(X>3)/E(X>3) which would just equal 1
I think I must be going about this problem all wrong but I don't know how to solve this, any help?
O wait, would it just be 1 minus the probability of being between 2 and 3? Or is that not right either?
Question seems weird. Expectation of X given X>3
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