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## 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} ##

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?

## \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?

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