Conditions on random variable to satisfy limit property

1. Sep 4, 2011

jjhyun90

1. The problem statement, all variables and given/known data
The problem is to find sufficient and preferably also necessary conditions on random variable X such that its characteristic function g(x) satisfies the limit property:
$\lim_{t\to0}\frac{1-g(\lambda t)}{1-g(t)}=\lambda^2$
I may assume X is symmetric around 0, so the characteristic function is real and even.

2. Relevant equations
$g(t)=\int_{-\infty}^{\infty} e^{itx}f_{X}(x) dx$

3. The attempt at a solution

I'm stuck immediately after trying to apply l'Hopital's rule. Any suggestions would be helpful.
Thank you.

2. Sep 4, 2011

Dickfore

What is the value of $g(0)$?

3. Sep 4, 2011

jjhyun90

$g(0)=1$ since it is a characteristic function.

4. Sep 4, 2011

Dickfore

So, what can you say about your limit? Can you evaluate it?

5. Sep 4, 2011

jjhyun90

Using l'Hopital's rule and chain rule, the equivalent condition I need will be
$\lim_{t\to0}\frac{g'(\lambda t)}{g'(t)} = \lambda$.

6. Sep 4, 2011

Dickfore

Not really, evaluate these derivatives explicitly:

$$\frac{d}{d t} \left(1 - g(\lambda t)\right)$$

$$\frac{d}{d t} \left(1 - g(t)\right)$$

7. Sep 4, 2011

jjhyun90

I must be confused. If I'm not mistaken, the first evaluates to $-\lambda g'(\lambda t)$ and the second $-g'(t)$.

8. Sep 4, 2011

Dickfore

$$\lim_{t \rightarrow 0}{\frac{-\lambda g'(\lambda t)}{-g'(t)}} = \lambda \, \lim_{t \rightarrow}{\frac{g'(\lambda t)}{g'(t)}}$$

Before you run in evaluating this limit, you need to consider two cases:

1) $g'(0) \neq 0$. Then the limit is simply 1 and you get the result you posted in post #5. However, this is not what you have in the condition of the problem.

2) $g'(0) = 0$ What is the limit in this case?

BTW, what does $g'(0)$ mean?

9. Sep 4, 2011

jjhyun90

The condition I posted on #5 is what I desire to have, rather than what will follow from the assumption. Sorry for the confusion.
By the assumption that X is symmetric around 0 and thus its expected value is 0, it follows g'(0)=0, so it is necessary to use l'Hopital's rule again.
Thus it seems a sufficient condition is that 0 < var(X) < infinity. Please let me know if I am wrong, but otherwise thank you for your help!

10. Sep 4, 2011

Dickfore

Yes, you are correct.