uniform moment generating function at t=0

Question

I have calculated the moment generating function for the uniform distribution as

$$M_X(t)=\frac{e^{tb}-e^{ta}}{t(b-a)}$$

However I know $M_X(0)=1$ but I can't get my head around how this is possible as if $t=0$, the denominator $= 0$ and therefore the function is undefined, where am I going wrong?

Answer 1

Technically speaking, $$ M_X(t)=\begin{cases}\displaystyle \frac{e^{tb}-e^{ta}}{t(b-a)} & \text{if $t\neq 0$}\\ 1 & \text{if $t=0$}\end{cases}. $$ If you look through the calculations that you used to come up with the MGF, you'll find that there's one point where you've divided by $t$... meaning that you would've needed to handle the case $t=0$ separately.

We don't tend to sweat this very much, however, because $M_X(t)\rightarrow1$ as $t\rightarrow0$ -- that is, your version of $M_X(t)$ has a removable discontinuity at $t=0$.

Answer 2

Recall that the moment generating function $M_X(t)$ is given by $$M_X(t)=E(e^{Xt}).$$ Put $t=0$. Then $e^{Xt}=e^0=1$. And of course $E(1)=1$, for any probability distribution.

Remark: A standard calculation shows that $$M_X(t)=\frac{e^{tb}-e^{ta}}{t(b-a)}$$ if $t\ne 0$.

It is not hard to verify, say using L'Hospital's Rule, that $$\lim_{t\to 0}\frac{e^{tb}-e^{ta}}{t(b-a)}=1.$$ Thus the function $\frac{e^{tb}-e^{ta}}{t(b-a)}$ has a removable singularity at $t=0$. The full moment generating function $M_X(t)$ is the only function which is $\frac{e^{tb}-e^{ta}}{t(b-a)}$ when $t\ne 0$ and which is continuous at $0$.

Answer 3

The numerator $e^{tb}-e^{ta}$ is also $0$ when $t=0$, so you should take a different approach, such as looking at the limit as $t \to 0$.

For small $t$, $\dfrac{e^{tb}-e^{ta}}{t(b-a)} \approx \dfrac{1+tb+t^2b^2/2 - 1-ta-t^2a^2/2}{t(b-a)} = 1 + \frac{b+a}{2}t$,

so it is reasonable for it to take the value $1$ when $t=0$

or using L'Hôpital's rule, $\lim \limits_{t \to 0}\dfrac{e^{tb}-e^{ta}}{t(b-a)} = \lim \limits_{t \to 0}\dfrac{be^{tb}-ae^{ta}}{b-a} = \dfrac{b-a}{b-a}=1$.