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Verify: If $0<\alpha<1$, $0\leq \beta$, then there exists $M(\alpha,\beta)>0$ so that $$z^\beta e^{-z}\leq Me^{-\alpha z}$$ for all $z\geq 0$

For part (1), I noticed that the left hand side of the inequality is equivalent to a gamma function. So, $$\int_0^\infty z^\beta e^{-z}\,dz=\beta !.$$ If I integrate the right hand side of the inequality I get $$\int_0^\infty M e^{-az}\,dz=M.$$ So if I choose $M(\alpha,\beta)\geq \beta!$. I have proven the inequality?

Is this correct?

EDIT: Proof: Let $f(z)=z^\beta e^{-(1-\alpha)z}$, with $0<\alpha<1$, $0\leq \beta$. Then $$f'(z) = \beta z^{\beta -1}e^{-(1-\alpha)z}+z^\beta e^{-(1-\alpha)z}(\alpha-1) = z^{\beta -1}e^{-(1-\alpha)z}[\beta +z(\alpha-1)] $$ Recall the first derivative test, which states:

"If $f'(x)>0$ on an open interval extending left from $x_0$ and $f'(x)<0$ on an open interval extending right from $x_0$, then $f(x)$ has a local maximum at $x_0$.

Note that $z^{\beta -1}\geq 0$ (since $z\geq 0$), $e^{-(1-\alpha)z}\geq 0$. So $f'(z)=0$ at $z=0$ and $z=-\dfrac{\beta}{\alpha-1}=\dfrac{\beta}{1-\alpha}$, with $z\geq 0$. When $z=0$ we can't test the open interval extending to the left, which doesn't give us any useful information.

When $z=\dfrac{\beta}{1-\alpha}$, choose $z_{-}=\dfrac{\beta-1}{1-\alpha}$ and $z_+=\dfrac{\beta +1}{1-\alpha}$. Then \begin{equation*} \begin{aligned} f'(z_{-}) & = \left(\dfrac{\beta-1}{1-\alpha}\right)^{\beta -1}\exp\left[-(1-\alpha)\dfrac{\beta-1}{1-\alpha}\right]\cdot \left[\beta +\dfrac{\beta-1}{1-\alpha}(\alpha-1)\right] \\ & = \left(\dfrac{\beta-1}{1-\alpha}\right)^{\beta -1}\exp\left[-(\beta-1) \right]\cdot 1 \\ f'(z_{+}) & = \left(\dfrac{\beta+1}{1-\alpha}\right)^{\beta -1}\exp\left[-(1-\alpha)\dfrac{\beta+1}{1-\alpha}\right]\cdot \left[\beta +\dfrac{\beta+1}{1-\alpha}(\alpha-1)\right] \\ & = \left(\dfrac{\beta+1}{1-\alpha}\right)^{\beta -1}\exp\left[-(\beta+1) \right]\cdot -1 \\ \end{aligned} \end{equation*}

However, it isn't obvious which is positive or negative. Help?

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  • $\begingroup$ As far as (a) goes that is about as far from a correct proof as you can get. Go back to basics and apply the first derivative test to $f(z) = z^\beta e^{(\alpha - 1)z}$. Worry about parts (b) and (c) later. $\endgroup$ – Umberto P. Feb 25 '18 at 21:40
  • $\begingroup$ I don't think it is helpful to say what you did about "As far as (a) goes that is about as far from a correct proof as you can get. Go back to basics..,". I find it very rude. But thank you for the hint. $\endgroup$ – Username Unknown Feb 25 '18 at 21:57
  • $\begingroup$ That's unfortunate, but I hope you understand your argument isn't fixable. You can report abusive posts with flags for moderator attention if you feel offended. $\endgroup$ – Umberto P. Feb 25 '18 at 22:09
  • $\begingroup$ @UmbertoP. I implemented your hint. So you can see the edits $\endgroup$ – Username Unknown Feb 25 '18 at 22:31
  • $\begingroup$ You are really close. Define $z_0 = \frac{\beta}{1-\alpha}$. You have $f'(z) > 0$ if $0 < z < z_0$ and $f'(z) < 0$ if $z > z_0$. What does this tell you? $\endgroup$ – Umberto P. Feb 25 '18 at 22:43

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