2
$\begingroup$

I would like to show that for $t > 0$, $$\int_{1}^{\infty}\frac{3e^{yt}}{y^4}\text{ d}y$$ diverges. [This is equivalent to showing that the MGF for $Y$, with pdf $$f_{Y}(y) = \dfrac{3}{y^4}\text{, } y \in (1, \infty)$$ does not exist.]


My work:

\begin{equation*} \int_{1}^{\infty}e^{yt} \cdot \dfrac{3}{y^4}\text{ d}y = \lim_{u \to \infty}3\int_{1}^{u}\dfrac{e^{yt}}{y^4}\text{ d}y\text{.} \end{equation*} Thus, it suffices to show that $\int_{1}^{\infty}\dfrac{e^{yt}}{y^4}\text{ d}y = \infty$. Recall that since $t > 0$ \begin{equation*} \lim_{y \to \infty}\dfrac{e^{yt}}{y^4} = \infty\text{.} \end{equation*} This means, by definition, that $\forall \alpha \in \mathbb{R}$, $\exists K > 0$ such that $\forall y > K$, \begin{equation*} \dfrac{e^{yt}}{y^4} > \alpha\text{.} \end{equation*} Thus, $\forall y > K$, obviously $\dfrac{e^{yt}}{y^4} > 1$. We may assume $K > 1$. Therefore, $\forall u \geq K$, \begin{align*} \int_{1}^{u}\dfrac{e^{yt}}{y^4}\text{ d}y &= \int_{1}^{K}\dfrac{e^{yt}}{y^4}\text{ d}y + \int_{K}^{u}\dfrac{e^{yt}}{y^4}\text{ d}y \\ &\geq \int_{1}^{K}\dfrac{e^{yt}}{y^4}\text{ d}y + \int_{K}^{u}1\text{ d}y \\ &\geq 0 + u - K \\ &= u - K\text{.} \end{align*} As $u \to \infty$, obviously $u - K \to \infty$.

Hence, $\displaystyle\int_{1}^{u}\dfrac{e^{yt}}{y^4}\text{ d}y \to \infty$ as well, and the integral diverges.


I'm just nervous about the sudden $K > 1$ assumption. Is this valid? Obviously $K$ must be quite large.

$\endgroup$
  • $\begingroup$ The integrand function is greater than $1$ for any $y$ large enough. $\endgroup$ – Jack D'Aurizio Oct 1 '16 at 15:07
  • $\begingroup$ Since exponential growth always wins over polynomials, the integrand $\to \infty$ at $\infty.$ $\endgroup$ – zhw. Oct 1 '16 at 15:11
2
$\begingroup$

Yes it is valid to take $K>1$.

We assume that $t>0$. One may observe that, for $K>1$, $$ \frac{3e^{yt}}{y^4}\ge \frac{3e^{yt}}{K^4}, \qquad y \in [1,K], $$ giving $$ \int_1^K\frac{3e^{yt}}{y^4}\:dy\ge \frac{3}{K^4}\int_1^K e^{ty}\:dy=\frac{3}{K^4}\cdot \frac{e^{K t}-e^t}{t} $$ the latter expression tends to $\infty$ as $K \to \infty$.

$\endgroup$

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.