Expected value of a variable and its reciprocal

[jeremy22511]

TL;DR Summary

If E(X1)>E(X2), does it mean that E(1/X1)<E(1/X2)? Thanks!

If E(X1)>E(X2), does it mean that E(1/X1)<E(1/X2)? Thanks!

 

[member 587159]

Hint: Try constant random variables.

 

[mathman]

No: Example two values X1=0 or 10. X2=1 or 3 (equal prob. for both).

E(X1)=5 and E(X2)=2. E(1/X1) is infinite, E(1/X2)=2/3.

 

[jeremy22511]

No: Example two values X1=0 or 10. X2=1 or 3 (equal prob. for both).

E(X1)=5 and E(X2)=2. E(1/X1) is infinite, E(1/X2)=2/3.

Thanks! What about the case with positive random variables? I probably should have given some context with my question. I was trying to make sense of using the two-tailed region in hypothesis testing by showing that it minimizes the expected value of P(H₀ | z-score in critical region). But this value depends on the expected value of the reciprocal of statistical power, which I can show to be maximized by the region. So I was trying to understand whether a higher E(beta) would translate to a lower E(1/beta).

 

[PeroK]

Summary:: If E(X1)>E(X2), does it mean that E(1/X1)<E(1/X2)? Thanks!

If E(X1)>E(X2), does it mean that E(1/X1)<E(1/X2)? Thanks!

Why don't you try some simple examples to see what happens? There seem to a lot of students who are unfamiliar with the concept of looking for a simple counterexample.

 

[member 587159]

No: Example two values X1=0 or 10. X2=1 or 3 (equal prob. for both).

E(X1)=5 and E(X2)=2. E(1/X1) is infinite, E(1/X2)=2/3.

$1/X_1$ makes absolutely no sense.

 

[Office_Shredder]

Instead of 0, try some arbitrarily small positive number.

 

[haruspex]

What about the case with positive random variables?

Try fixing Y as 1 and letting X take two values, equally likely, with a mean slightly greater than 1.

 

[PeroK]

In general, if $X$ takes two values $a, b$ with equal probability, then $E(\frac 1 X) = \frac 1 {E(X)}$ only when $a = b$:
$$E(\frac 1 X) = \frac 1 2 (\frac 1 a + \frac 1 b) = \frac 1 {ab}(\frac{a + b}{2}) = \frac 1 {ab} E(X)$$
$$E(\frac 1 X) = \frac 1 {E(X)} \ \Rightarrow \ E(X)^2 = ab \ \Rightarrow \ (a+b)^2 = 4ab \ \Rightarrow \ (a-b)^2 = 0$$

 

[BWV]

Look at a uniform dist X on (0.01,0.99). E(X) =0.5 and Y a uniform on (0.35,0.45) so E(Y)=0.4

so E(X)>E(Y), but also E(1/X)>E(1/Y) as E(1/Y) =2.5 and E(1/X)=4.7

 

[haruspex]

In general, if $X$ takes two values $a, b$ with equal probability, then $E(\frac 1 X) = \frac 1 {E(X)}$ only when $a = b$:
$$E(\frac 1 X) = \frac 1 2 (\frac 1 a + \frac 1 b) = \frac 1 {ab}(\frac{a + b}{2}) = \frac 1 {ab} E(X)$$
$$E(\frac 1 X) = \frac 1 {E(X)} \ \Rightarrow \ E(X)^2 = ab \ \Rightarrow \ (a+b)^2 = 4ab \ \Rightarrow \ (a-b)^2 = 0$$

If that's in reply to post #8, I was not suggesting 1/E(X)=E(1/X). The idea is to find a pair of values for X such that E(X)>1 =E(Y) and E(1/X)>1=E(1/Y).
@BWV seems to have come up with a closely related example.

 

[PeroK]

If that's in reply to post #8, I was not suggesting 1/E(X)=E(1/X). The idea is to find a pair of values for X such that E(X)>1 =E(Y) and E(1/X)>1=E(1/Y).
@BWV seems to have come up with a closely related example.

It was just a general post to show that one doesn't have to be clever to find an example. The result the OP was looking for can only be true in a special case.

 

[haruspex]

It was just a general post to show that one doesn't have to be clever to find an example. The result the OP was looking for can only be true in a special case.

Hmm... ok, but I don't see how your post shows that. How does it give an example of E(X)>E(Y) while E(1/X)>E(1/Y)?

 

[PeroK]

Hmm... ok, but I don't see how your post shows that. How does it give an example of E(X)>E(Y) while E(1/X)>E(1/Y)?

I forgot that bit!

 

[PeroK]

To make amends for post #9. If we let $a > 1$ and consider $X$ equally likely to be $a$ or $\frac 1 a$, then:
$$E(X) = E(\frac 1 X) = \frac 1 2 (a + \frac 1 a)$$
Hence, as $a$ increases both $E(X)$ and $E(\frac 1 X)$ increase.

 

[BWV]

More challenging to find an example where E(X)>E(Y) and E(1/X) >E(1/Y)
where Var(X) < Var(Y).

 

[FactChecker]

More challenging to find an example where E(X)>E(Y) and E(1/X) >E(1/Y)
where Var(X) < Var(Y).

Not really hard. Let X== 1 and Y be uniform on [-2,-1].

 

[haruspex]

Not really hard. Let X== 1 and Y be uniform on [-2,-1].

I think it was sort of agreed earlier in the thread that X and Y should be taken as always nonnegative.

 

[FactChecker]

I think it was sort of agreed earlier in the thread that X and Y should be taken as always nonnegative.

In that case, I suspect it is impossible. I am not sure how to prove it.

 

[BWV]

actually for positive numbers, just need one number close to zero with a small weight in the distribution
for example:
X=[10^-20,1000:2000]
Y=1:1999

E(X)=1499 >E(Y)=1000
Var(X)=8.6*10^4<Var(Y)=3.3*10^5

E(X^-1)=10^16>E(Y^-1)<1