Conditions on random variable to satisfy limit property

[jjhyun90]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

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

 

[Dickfore]

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

 

[jjhyun90]

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

 

[Dickfore]

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

 

[jjhyun90]

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

 

[Dickfore]

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

Not really, evaluate these derivatives explicitly:

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

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

 

[jjhyun90]

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

 

[Dickfore]

True, so your limit becomes:

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

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?

 

[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!

 

[Dickfore]

Yes, you are correct.