HW question on using Chebyshev's inequality with a moment generating function

Question

I'm not sure how to exactly approach this problem if anyone could give any advice that would be great thank you.

Answer 1

Since $m_Y^\prime(t)=m_Y(t)10t\exp\tfrac{t^2}{2}$,$$m_Y^{\prime\prime}(t)=m_Y(t)(100t^2\exp t^2+10(1+t^2)\exp\tfrac{t^2}{2}).$$Thus$$m_Y(0)=1,\,\Bbb EY=m_Y^\prime(0)=0,\,\Bbb EY^2=m_Y^{\prime\prime}(0)=10\implies\operatorname{Var}Y=10.$$Now you can use the inequality.

Answer 2

What if I gave your something (provably!) better than what you can get with Chebyshev, by using all the information provided by the characteristic funciton of your random variable $Y$ ?

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For every $\epsilon,t>0$, we have $$ \mathbb P(Y > \epsilon) = \mathbb P(tY > t\epsilon) = \mathbb P(e^{tY} > e^{t\epsilon}) \le \frac{\mathbb E[e^{tY}]}{e^{t\epsilon}} = \frac{m_{Y}(t)}{e^{t\epsilon}}, $$ where the last inequality is an application of Markov's inequality. Thus, we have $$ \mathbb P(Y > \epsilon) \le \min_{t > 0}e^{-t\epsilon}m_{Y}(t) =: h(\epsilon). $$ (Do the computation yourself to determine the function $h$).

Similarly, because $m_Y$ is odd (see question description), we have

$$ \mathbb P(Y < -\epsilon) = \mathbb P(-Y > \epsilon) \le \min_{t > 0}e^{-t\epsilon}m_{-Y}(t) = \min_{t > 0}e^{-t\epsilon}m_{-Y}(t) = \min_{t > 0}e^{-t\epsilon}m_{Y}(t)=h(\epsilon). $$

Putting things together, we get

$\mathbb P(|Y| > \epsilon) = \mathbb P(Y > \epsilon) + \mathbb P(Y < -\epsilon) \le h(\epsilon) + h(\epsilon) = 2h(\epsilon)$.

Now plug-in $\epsilon = 4$ to solve your specific problem.