Is it even possible to find the variance of this moment generating function?

Question

This is my moment generating function: $M_x(t) = \frac{6e^t}{t^2} + \frac{6}{t^2} + \frac{12e^t}{t} - \frac{12e^t}{t^3} + \frac{12}{t^3}$.

I have to find the mean the variance of it.

After taking the first derivative ($M_x'(t) = -\frac{18e^t}{t^2} - \frac{12}{t^3} + \frac{12e^t}{t} + \frac{36e^t}{t^4} - \frac{36}{t^4}$) and using L'Hopital's rule over and over again, my mean is $E(X) = \frac{1}{2}$.

But I am having problems finding the variance of this function and not sure if it even exists. This is my second derivative for $M_x(t)$:

$$M_x''(t)= \frac{-30e^t t^3 + 36e^tt^2 + 36t - 12e^tt^4 + 36e^tt - 144e^t + 144}{t^5}$$

I can only apply L'Hopital's rule once as the limit of t goes to $0$, and that's the problem as I want to use L'Hopital's rule over and over again to get $E(X^2)$ for my variance.

So is there a trick you guys think I'm missing?

Answer 1

The question is a mess and the proposed function is not an MGF. Fortunately, from what is hidden somewhere in the comments, one can deduce that the question is as follows:

Consider $(X_1,X_2)$ with density $f(x_1,x_2)=12x_1x_2(1-x_2)$ on $0\lt x_1,x_2\lt1$. Find the MGF of $X_1$ and $X_2$ and deduce their means and variances.

Well, if I had to solve this, I would first notice that the joint density can be factored hence $X_1$ and $X_2$ are independent with respective densities $f_1$ and $f_2$ with $$f_1(x)=2x\mathbf 1_{0\lt x\lt1},\qquad f_2(x)=6x(1-x)\mathbf 1_{0\lt x\lt1}.$$ Thus, for example, the MGF $M_1$ of $X_1$ is such that, for every $t$, $$M_1(t)=E(\mathrm e^{tX_1})=\int_0^1\mathrm e^{tx}2x\mathrm dx=\left.2t^{-2}(tx-1)\mathrm e^{tx}\right|_0^1=2t^{-2}(1-(1-t)\mathrm e^t).$$ To deduce the mean and variance, the simplest approach might be to expand the exponential $\mathrm e^t$ in $M_1(t)$ up to fourth order and to collect the $t$ and $t^2$ terms in $M_1(t)$.

And similarly for $M_2(t)=E(\mathrm e^{tX_2})$.

Answer 2

There's definitely something wrong with your MGF: it has a singularity as $t \to 0$. In fact $$M(t) = -12 t^{-2} + 7 + \dfrac{9}{2} t + \ldots$$ as can be shown most easily using Taylor series. So it is not a moment generating function.