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My pdf is defined as follows:

$$f_X(x) = \frac{1}{\tau} e^{-x/\tau}$$

At first I started finding the characteristic function like so:

$$\hat{f}_X(\xi) = \mathbb{E}[e^{i\xi X}] = \frac{1}{\tau}\int_{\mathbb{R}} e^{i\xi x}e^{-x/\tau}dx$$

I then wrote $e^{i\xi x}$ as $\cos\xi x + i \sin \xi x$, so that I have:

$$\frac{1}{\tau} \int_{\mathbb{R}}\cos (\xi x) e^{-x/\tau} dx + \frac{i}{\tau} \int_{\mathbb{R}}\sin(\xi x)e^{-x/\tau}dx$$

Which in hindsight is not the best of the ideas (even though I can find the integrals, I will have problems evaluating say $\cos(\xi x)$ at infinity. So how shall I actually solve this?

Naz
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    First the integrals are on $(0,+\infty)$, not $\mathbb R$. Second, one should know a primitive of the function $x\mapsto e^{i\xi x}e^{-x/\tau}=e^{zx}$ for $z=i\xi-1/\tau$. – Did Apr 08 '16 at 10:43
  • got it. thanks $ $ – Naz Apr 08 '16 at 10:45
  • @Did wait why is the lower bound $0$. The definition for characteristic function has minus infinity. The pdf that I have exists in the negative domain as well – Naz Apr 08 '16 at 11:00
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    the PDF you have is non-negative only on $x \geq 0$; otherwise, it can't integrate to $1$ (in fact, the PDF blows up as $x\to -\infty$ if that was for all of $\mathbb{R}$ so it couldn't integrate to $1$). – Batman Apr 08 '16 at 11:11
  • Yes the PDF exists on the whole real line -- and this whole exchange is yet another indication that probability density functions should always be defined on the whole real line. In your case, for every real $x$, $$f_X(x)=\frac{1}{\tau} e^{-x/\tau}\mathbf 1_{x>0},$$ hence, for every real $\xi$, $$\hat{f}X(\xi) = \int{\mathbb R}e^{i\xi x}f_X(x)dx= \int_{\mathbb R}e^{i\xi x}\frac{1}{\tau}e^{-x/\tau}\mathbf 1_{x>0}dx=\frac{1}{\tau}\int_{0}^\infty e^{i\xi x}e^{-x/\tau}dx.$$ – Did Apr 08 '16 at 12:40
  • @isquared-KeepitReal Did you find the characteristic function? – JKnecht Apr 08 '16 at 14:35
  • I am going to now. I had to obtain my moments since I wasn't using the correct bounds. – Naz Apr 08 '16 at 14:49
  • @JKnecht is it $\frac{1}{1+i \tau \xi}$? (I had to multiply $i$ by infinity in the bounds though...); I noticed it says that you were a trader on your page :) I am doing a financial maths course right now :) – Naz Apr 08 '16 at 15:19
  • I dont know, i havent done the calculation. But i can do it later today and post an answer. Yes, i was. I can recommend it. Its good fun :) especially when you trade with your own money :) But being back at the university, without any pressure, that is even better. – JKnecht Apr 08 '16 at 15:32

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On request i checked if OP got the right CF. Almost. Just the wrong sign.

$$ \begin{align} \hat{f}_X(\xi) & = \mathbb{E}[e^{i\xi X}] \\ & \\ & = \frac{1}{\tau}\int_{0}^{\infty} e^{i\xi x}e^{-x/\tau}\:\:\text{dx}\\ & \\ & = \frac{1}{\tau}\int_{0}^{\infty} e^{-(1/\tau - i\xi) x}\:\:\text{dx}\\ & \\ & = -\frac{1}{\tau}\frac{1}{1/\tau - i\xi} \left[e^{-(1/\tau - i\xi) x}\:\right]_0^{\infty}\\ & \\ & = \frac{1}{\tau}\frac{1}{1/\tau - i\xi} \\ & \\ & = \frac{1}{1 - i\tau\xi} \\ \end{align} $$

JKnecht
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  • I suspected the sign was wrong because I was getting incorrect expectation values. Thanks ;) – Naz Apr 09 '16 at 05:14