Expected Value of Random Variable X: Solving for E[1/X]

[Steve Zissou]

Hello all,
I'm wondering if someone can offer some insight here: We have a random variable X, and it's expectation is called y.
Can it be shown that
1/y = E[1/X]
??
Thanks

 

[mathman]

Not usually. Example: coin flip with heads (X=0) and tails (X=1). The expectation is 1/2. The expectation of 1/X is infinite.

 

[Steve Zissou]

Thanks for your quick reply, mathman.
I should have specified X is never zero.

 

[Dale]

Hello all,
I'm wondering if someone can offer some insight here: We have a random variable X, and it's expectation is called y.
Can it be shown that
1/y = E[1/X]
??
Thanks

Definitely not. Consider the random variable that takes values 1 and -1 with equal probability. The expectation is y=0 so 1/y is undefined. But E[1/X] is 0.

 

[Steve Zissou]

Thanks Dale, as mentioned, I should have specified X is never zero, in fact it is always positive, and hence y>0.
Thanks for your reply though!

 

[Dale]

It doesn't really matter, the point it that the relationship doesn't hold in general. That was just the easiest counterexample I could come up with in my head. Take pretty much any pair of numbers and you will get similar results.

 

[Steve Zissou]

Dale:
Right, I see what you're saying. If X={1,2,3} the y = 2. But then E[1/X] = 11/18.
Could perhaps we say in general if y = E[X] that maybe 1/y < E[1/X] or perhaps even 1/y =< E[1/X] ?
Thanks

 

[Steve Zissou]

Wait a second. This is Jensen's Inequality: f(E[X])=<E[f(X)]
In my case, we have f(X) = 1/X and y = E[X]. So I can say
1/y =<E[1/X].

 

[Dale]

That could be. I think that 1/x is a convex function.