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In Casella's statistical inference textbook exercise 3.21, it asked the proof to show Cauchy distribution doesn't have mgf. One solution I find online is:

The moment generating function would be defined by $\frac{1}{\pi} \int_{-\infty}^\infty \frac{e^{tx}}{1+x^2} dx$. On $(0, \infty)$, $e^{tx} > x$, hence

$$\int_0^\infty \frac{e^{tx}}{1+x^2} dx > \int_0^\infty \frac{x}{1+x^2} dx =\infty$$

thus the moment generating function does not exist.

I don't think this proof is rigorous. First, I am not convinced by "On $(0, \infty)$, $e^{tx} > x$". Second, the integral only considers the $(0, \infty)$ part.

Jackie
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Moment generating functions always exist, as long as you are fine with infinite values. Here we show that for Cauchy distribution mgf is always infinite, when $t \neq 0$.

Inequality $\mathrm{e}^{tx} > x$ holds for any positive $t$ and large enough $x$, which is enough to determine divergence. For example, if it held for $x > x_0$, then you could change bounds of integration to $(x_0, \infty)$.

You are missing one inequality: $$ \int_{-\infty}^\infty \frac{e^{tx}}{1+x^2} dx > \int_{x_0}^\infty \frac{e^{tx}}{1+x^2} dx $$ It holds, because function we are integrating is always positive. If you tried negative $t$, you could make the same argument on $(-\infty, 0)$ instead.

But really, divergence of $\int_{x_0}^\infty f(x) dx$ implies divergence of $\int_{-\infty}^\infty f(x) dx$ by definition of improper integral with multiple singularities, so last inequality isn't needed.

Esgeriath
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