I know that $\phi_Y(t) = E(e^{itY})=E(\cos(tY))+iE(\sin(tY))$

After integration I have found that $E(\cos(tY))= \frac{\sin(t)}{t}$ and $E(\sin(tY))=0$. So is the characteristic function just $\frac{\sin(t)}{t}$?

  • 3
    $\begingroup$ This calculation is illustrated for the $U(0,1)$ distribution near the beginning of my post at stats.stackexchange.com/a/43075. Comparing your approach to that one might increase your confidence in your result. $\endgroup$
    – whuber
    Oct 20, 2015 at 19:33

1 Answer 1


Here is another way to approach this. The characteristic function might be viewed as the moment generating function evaluated at $it$, in case the latter exists of course. But it does for this distribution (a bounded distribution you see), so let's find it. We have

$$\psi (t) = \int_{-1}^1 \frac{e^{tx}}{2}dx =\frac{1}{2t} \left[e^{t}- e^{-t} \right]$$

And evaluating at $it$ we get,

$$\begin{align} \psi (it)=\frac{1}{2it} \left[ e^{it}-e^{-it} \right] &= \frac{1}{2it}\left[ \cos t + i \sin t -\left( \cos t- i\sin t \right) \right] \\ &= \frac{1}{2it} \left[ 2i \sin t\right] \\ &= \frac{\sin t}{t} \end{align} $$

And you were right of course. Exploiting the connection with the moment generating function can often provide a shortcut and also a way to verify your results.

Hope this helps.


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