What is the nth derivative of this equation?

[Neothilic]

TL;DR

What is the solution to the nth derivative of this characteristic function?

How would I find the $nth$ derivative of this? As as derivative of $-|t|$ is $-\frac{t}{|t|}$.
$\langle X^{n} \rangle = i^{-n}\frac{d^{n}}{dt^n} e^{-|t|} \vert_{t=0}$

This is the characteristic function of the Cauchy Distribution. So for when $t=0$, $e^{-|t|}=1$; when $t<0$, $e^{-|t|}=e^{t}$; when $t>0$, $e^{-|t|}=e^{-t}$. So I could just substitute the value for when $t=0$. Correct?

As;
$\langle X^{n} \rangle = i^{-n}\frac{d^{n}}{dt^n} e^{-|t|} \vert_{t=0} = i^{-n}\frac{d^{n}}{dt^n} 1$

Therefore;
$\langle X^{n} \rangle = i^{-n}\frac{d^{n}}{dt^n} e^{-|t|} \vert_{t=0}=i^{-n}0$

So;
$\langle X^{n} \rangle = i^{-n}\frac{d^{n}}{dt^n} e^{-|t|} \vert_{t=0}=0$

Therefore, the Cauchy Distribution has no moments.

I know the result is true, is my solution correct though?

 

[Haborix]

There is a difference between the moments being zero and them not being well-defined; you make both claims. You start out in the right direction by noting the absolute value makes things more difficult. However, your calculation is not correct. The evaluation at $t=0$ comes after differentiation. Have you tried sketching the characteristic function and looking at its behavior in the region of $t=0$?

 

[Neothilic]

There is a difference between the moments being zero and them not being well-defined; you make both claims. You start out in the right direction by noting the absolute value makes things more difficult. However, your calculation is not correct. The evaluation at $t=0$ comes after differentiation. Have you tried sketching the characteristic function and looking at its behavior in the region of $t=0$?

So I sketches graph, and it is discontinuous at $t=0$. This means it is not differentiable for any $n$. As $C(t)$ is not differentiable for any $n$, and it is part of the equation for the moments of the Cauchy Distribution, the Cauchy Distribution has no moments that exist for any $n$. Is that better?

 

[Haborix]

That's the idea. If you were inclined to try and formalize it in equations, you could take the derivative of the function on either side of zero and show the discontinuity as you took the limit from the right and the left.

 

[Neothilic]

That's the idea. If you were inclined to try and formalize it in equations, you could take the derivative of the function on either side of zero and show the discontinuity as you took the limit from the right and the left.

I have done some research and have gotten to this point;

$\lim_{t \to 0} C(t)= \frac{C(t_0)-C(t)}{t_0-t}$

$\lim_{t \to 0^{+}} C(t)= \frac{C(0^{+})-C(t)}{0^{+}-t}=\frac{1-e^{-|t|}}{t} (1)$

$\lim_{t \to 0^{-}} C(t)= \frac{C(0^{-})-C(t)}{0^{-}-t}=\frac{1-e^{-|t|}}{-t} (2)$

As $(1)$ does not equate to $(2)$, $C(t)$ is discontinuous. Is this correct?

 

[PeroK]

I have done some research and have gotten to this point;

$\lim_{t \to 0} C(t)= \frac{C(t_0)-C(t)}{t_0-t}$

$\lim_{t \to 0^{+}} C(t)= \frac{C(0^{+})-C(t)}{0^{+}-t}=\frac{1-e^{-|t|}}{t} (1)$

$\lim_{t \to 0^{-}} C(t)= \frac{C(0^{-})-C(t)}{0^{-}-t}=\frac{1-e^{-|t|}}{-t} (2)$

As $(1)$ does not equate to $(2)$, $C(t)$ is discontinuous. Is this correct?

Why not use:
$$e^{-|t|} = e^{-t} \ (t > 0) \ \ \text{and} \ \ e^{-|t|} = e^{t} \ (t < 0)$$

 

[Neothilic]

Why not use:
$$e^{-|t|} = e^{-t} \ (t > 0) \ \ \text{and} \ \ e^{-|t|} = e^{t} \ (t < 0)$$

That's a great point!

$\lim_{t \to 0} C(t)= \frac{C(t_0)-C(t)}{t_0-t}$

$\lim_{t \to 0^{+}} C(t)= \frac{C(0^{+})-C(t)}{0^{+}-t}=\frac{1-e^{-t}}{t} (1)$

$\lim_{t \to 0^{-}} C(t)= \frac{C(0^{-})-C(t)}{0^{-}-t}=\frac{1-e^{t}}{t} (2)$

That looks better?

 

[PeroK]

That's a great point!

$\lim_{t \to 0} C(t)= \frac{C(t_0)-C(t)}{t_0-t}$

$\lim_{t \to 0^{+}} C(t)= \frac{C(0^{+})-C(t)}{0^{+}-t}=\frac{1-e^{-t}}{t} (1)$

$\lim_{t \to 0^{-}} C(t)= \frac{C(0^{-})-C(t)}{0^{-}-t}=\frac{1-e^{t}}{t} (2)$

That looks better?

The original function $e^{-|t|}$ is continuous but not differentiable at $t = 0$. The best you can do is differentiate it separately for $t > 0$ and $t < 0$. You don't need to mess about with limits for this.

I must admit I don't know what you are supposed to do to calculate the moments. Technically, only the zeroth moment exists. Beyond that the nth derivatives do not exist. But, for the even derivatives you have equal limits as $t \rightarrow 0^{\pm}$. Perhaps the even moments are taken to exist in that case? I.e. if you take the limit $t \rightarrow 0$ instead of "evaluated at $t = 0$".

 

[PeroK]

Summary:: What is the solution to the nth derivative of this characteristic function?

How would I find the $nth$ derivative of this? As as derivative of $-|t|$ is $-\frac{t}{|t|}$.
$\langle X^{n} \rangle = i^{-n}\frac{d^{n}}{dt^n} e^{-|t|} \vert_{t=0}$

Technically, these derivatives are not defined at $t = 0$. But, if you calculate instead:
$$\langle X^{n} \rangle = i^{-n}\lim_{t \rightarrow 0} \frac{d^{n}}{dt^n} e^{-|t|}$$
Then the even moments exist.

 

[Neothilic]

The original function $e^{-|t|}$ is continuous but not differentiable at $t = 0$. The best you can do is differentiate it separately for $t > 0$ and $t < 0$. You don't need to mess about with limits for this.

I must admit I don't know what you are supposed to do to calculate the moments. Technically, only the zeroth moment exists. Beyond that the nth derivatives do not exist. But, for the even derivatives you have equal limits as $t \rightarrow 0^{\pm}$. Perhaps the even moments are taken to exist in that case? I.e. if you take the limit $t \rightarrow 0$ instead of "evaluated at $t = 0$".

What I want to do, is to prove the Cauchy Distribution has no moments. To do this, I want to use the characteristic function. So firstly, I have drawn the graph of the characteristic function to show it is discontinuous at $t=0$ so meaning it cannot be differentiated at $t=0$. Now I want to prove this using actually maths instead of intuitively. Which is why I was trying to show it using limits.

As you have correctly said, $C(t)$ is not differentiable at $t=0$. So I want to show that mathematically, instead of just stating it.

Edit: I think I have found a video, explaining what it is I need to do.

 

[PeroK]

How much real analysis do you know?

 

[PeroK]

That's a great point!

$\lim_{t \to 0} C(t)= \frac{C(t_0)-C(t)}{t_0-t}$

$\lim_{t \to 0^{+}} C(t)= \frac{C(0^{+})-C(t)}{0^{+}-t}=\frac{1-e^{-t}}{t} (1)$

$\lim_{t \to 0^{-}} C(t)= \frac{C(0^{-})-C(t)}{0^{-}-t}=\frac{1-e^{t}}{t} (2)$

That looks better?

These don't look right.
For $C$ to be differentiable at $0$ you need both limits to exist and:
$$\lim_{t \to 0^{+}} \frac{C(t)-C(0)}{t} = \lim_{t \to 0^-} \frac{C(t)-C(0)}{t}$$
Where, for example:
$$ \lim_{t \to 0^{+}} \frac{C(t)-C(0)}{t}=\lim_{t \to 0^{+}}\frac{e^{-t}- 1}{t}$$
If you want a quick proof, then use a Taylor series.

 

[Neothilic]

These don't look right.
For $C$ to be differentiable at $0$ you need both limits to exist and:
$$\lim_{t \to 0^{+}} \frac{C(t)-C(0)}{t} = \lim_{t \to 0^-} \frac{C(t)-C(0)}{t}$$
Where, for example:
$$ \lim_{t \to 0^{+}} \frac{C(t)-C(0)}{t}=\lim_{t \to 0^{+}}\frac{e^{-t}- 1}{t}$$
If you want a quick proof, then use a Taylor series.

I agree it is wrong.

 

[Neothilic]

How much real analysis do you know?

I have studied a module on it at university a couple years ago, but I can't remember a lot off hand though.

 

[PeroK]

I have studied a module on it at university a couple years ago, but I can't remember a lot off hand though.

Just use Taylor series. Or draw a graph of the function!

 

[Neothilic]

These don't look right.
For $C$ to be differentiable at $0$ you need both limits to exist and:
$$\lim_{t \to 0^{+}} \frac{C(t)-C(0)}{t} = \lim_{t \to 0^-} \frac{C(t)-C(0)}{t}$$
Where, for example:
$$ \lim_{t \to 0^{+}} \frac{C(t)-C(0)}{t}=\lim_{t \to 0^{+}}\frac{e^{-t}- 1}{t}$$
If you want a quick proof, then use a Taylor series.

Let us try and find the derivative of $e^{-|t|}$ at $t=0$. As we know;

$\frac{d}{dy}(e^{y})=\frac{d}{dy}(y)e^{y}$

Therefore;

$\frac{d}{dy}(e^{-|t|})=\frac{d}{dy}(-|t|)e^{-|t|}$

Let us try and find;

$\frac{d}{dy}(-|t|)$

$\frac{d}{dy}(-|t|)=-\lim_{\delta t \to 0} \frac{(|0+\delta t)+0|)-|0|}{\delta t}=\lim_{\delta t \to 0} \frac{(-|t|)}{\delta t}$

$\lim_{\delta t \to 0^{-}} \frac{(-|t|)}{\delta t}=-\frac{(|-NUMBER|)}{-NUMBER}=\frac{NUMBER}{NUMBER}=1 (1)$

$\lim_{\delta t \to 0^{+}} \frac{(-|t|)}{\delta t}=-\frac{(|NUMBER|)}{NUMBER}=-\frac{NUMBER}{NUMBER}=-1 (2)$

As $(1)$ does not equal $(2)$, $-|t|$ is not differentiable at $t=0$, which means $e^{-|t|}$ is not differentiable at $t=0$.

 

[PeroK]

Here is the approach I intended. Let $f(t) = e^{-|t|}$. For $t < 0$ we have $f(t) = e^t = 1 + t + \frac{t^2}{2!} \dots$, and for $t > 0$ we have $f(t) = e^{-t} = 1 - t + \frac{t^2}{2!}$. Hence:
$$\lim_{t \rightarrow 0^-}\frac{f(t) - f(0)}{t} = \lim_{t \rightarrow 0^-}\frac{t + \frac{t^2}{2!} + \dots}{t} = 1$$
$$\lim_{t \rightarrow 0^+}\frac{f(t) - f(0)}{t} = \lim_{t \rightarrow 0^+}\frac{-t + \frac{t^2}{2!} + \dots}{t} = -1$$
The right and left limits are different at $t=0$, hence the function is not differentiable at $t = 0$.

 

[Neothilic]

Here is the approach I intended. Let $f(t) = e^{-|t|}$. For $t < 0$ we have $f(t) = e^t = 1 + t + \frac{t^2}{2!} \dots$, and for $t > 0$ we have $f(t) = e^{-t} = 1 - t + \frac{t^2}{2!}$. Hence:
$$\lim_{t \rightarrow 0^-}\frac{f(t) - f(0)}{t} = \lim_{t \rightarrow 0^-}\frac{t + \frac{t^2}{2!} + \dots}{t} = 1$$
$$\lim_{t \rightarrow 0^+}\frac{f(t) - f(0)}{t} = \lim_{t \rightarrow 0^+}\frac{-t + \frac{t^2}{2!} + \dots}{t} = -1$$
The right and left limits are different at $t=0$, hence the function is not differentiable at $t = 0$.

I see, well this way is more efficient than the way I proposed. I also read on wiki, the reason why the Cauchy Lorentz distribution has no moment is because the characteristic function has no Taylor expansion about $t=0$. I am skeptical about you way, but you seem confident.

 

[PeroK]

I see, well this way is more efficient than the way I proposed. I also read on wiki, the reason why the Cauchy Lorentz distribution has no moment is because the characteristic function has no Taylor expansion about $t=0$. I am skeptical about you way, but you seem confident.

I didn't Taylor expand $f$ about $t = 0$. I took the Taylor expansion of $e^t$ and $e^{-t}$ in the separate regions where these are valid representations of $f$.

The fact that these Taylor expansion are different for $t > 0$ and $t < 0$ shows that $f$ has no Taylor expansion at $t =0$.

 

[Neothilic]

I didn't Taylor expand $f$ about $t = 0$. I took the Taylor expansion of $e^t$ and $e^{-t}$ in the separate regions where these are valid representations of $f$.

The fact that these Taylor expansion are different for $t > 0$ and $t < 0$ shows that $f$ has no Taylor expansion at $t =0$.

Yes, my mistake. I should have re-evaluated my reply. I appreciate the help. So, do you think the solution you proposed has more validity than the solution I proposed?

 

[PeroK]

Yes, my mistake. I should have re-evaluated my reply. I appreciate the help. So, do you think the solution you proposed has more validity than the solution I proposed?

I don't really understand quite what you did!

 

[Neothilic]

I don't really understand quite what you did!

The video that I linked earlier in the thread, I followed exactly what that teacher did!

 

[PeroK]

The video that I linked earlier in the thread, I followed exactly what that teacher did!

He's not doing an exponential. For example:

$\frac{d}{dy}(e^{-|t|})=\frac{d}{dy}(-|t|)e^{-|t|}$

Leaving aside the confusion between $y$ and $t$, this is not valid for $t = 0$. You can't differentiate $e^{-|t|}$ at $t = 0$, when you are trying to prove it's not differentiable at that point!

What you could have done was:

If $e^{-|t|}$ is differentiable at $t = 0$, then $\ln(e^{-|t|}) = -|t|$ must be differentiable at $t = 0$ (as $\ln$ is differentiable at $1$). But $-|t|$ is not differentiable at $t = 0$, hence neither is $e^{-|t|}$. That would be a neat way to prove it.

 

[Neothilic]

He's not doing an exponential. For example:
Leaving aside the confusion between $y$ and $t$, this is not valid for $t = 0$. You can't differentiate $e^{-|t|}$ at $t = 0$, when you are trying to prove it's not differentiable at that point!

What you could have done was:

If $e^{-|t|}$ is differentiable at $t = 0$, then $\ln(e^{-|t|}) = -|t|$ must be differentiable at $t = 0$ (as $\ln$ is differentiable at $1$). But $-|t|$ is not differentiable at $t = 0$, hence neither is $e^{-|t|}$. That would be a neat way to prove it.

I understand what you mean that I can't show it's not differentiable in the way I did it. I think the way you did it previously is suffice. As I like the use of $e^{-t}$ and $e^{t}$.