0
$\begingroup$

I am trying to use the characteristic function of the uniform distribution defined on (0,1) to compute the mean. I have calculated the characteristic function (correctly) and used Euler's identity to convert it to the following form:

$$\phi_Y(t)=\frac{\sin(t)}{t} + i \frac{1-\cos(t)}{t}$$

I should be able to compute the mean (which should be 1/2) by taking the first derivative, multiplying by $\frac{1}{i}$, and evaluating at $t=0$. I've computed the first derivative as:

$$\frac{\partial}{\partial t}\phi_Y(t)=\frac{t\cos(t)-\sin(t)}{t^2} + i\frac{t \sin(t) + \cos(t) -1}{t^2}$$

And dividing by $i$, this expression simplifies to: $$E[X]=\Big(\frac{i\sin(t)-it\cos(t)+t(\sin(t)+\cos(t)-1}{t^2}\Big)\bigg\rvert_{t=0}$$

This expression is undefined, because of division by 0. Am I missing something here?

$\endgroup$
  • $\begingroup$ You cannot just "plug in $t=0$" indeed; but the limit as $t\to 0$ does exist. we have $\lim_{t\to 0}\frac{\sin t}{t} = 1$, $\lim_{t\to 0}\frac{\cos t-1}{t^2} = -1/2$, etc. (You can do a Taylor expansion of the numerator at $t=0$, if you are familair with Taylor expansions; this will be immediate) $\endgroup$ – Clement C. Sep 11 '18 at 5:19
  • $\begingroup$ Oh, I see. Thank you very much! $\endgroup$ – Austin D. Sep 11 '18 at 16:18
0
$\begingroup$

Use Maclaurin series: $$i\sin t-it\cos t+t\sin t+\cos t-1 =it-it+t^1+1-t^2/2-1)+O(t^3)=\frac{t^2}2+O(t^3)$$ and so $$\lim_{t\to0}\frac{i\sin t-it\cos t+t\sin t+\cos t-1 }{t^2}=\frac12.$$

$\endgroup$
  • $\begingroup$ Oh, I see. Thank you very much! $\endgroup$ – Austin D. Sep 11 '18 at 16:18
0
$\begingroup$

You have to use limiting values. Even in your formula for $\phi_Y (t)$ there is $t$ in the denominator. It doesn't mean $\phi_Y (0)$ is not defined. The value is $1$ which is also $\lim_{t \to 0} \phi_Y (t)$. Similarly, you can find $EY$ by computing $\lim _{t \to 0} \frac {\partial} {\partial t} \phi_Y (t)$. By expanding $\sin $ and $cos $ in their Taylor series you get the value of the limit as $\frac 1 2$.

$\endgroup$
  • $\begingroup$ Oh, I see. Thank you very much! $\endgroup$ – Austin D. Sep 11 '18 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.