A) Prove Bernstein's inequality: If X is a random variable for which Mx(t) exists for some t>0, the P(X>=x) <= Mx(t)e^(-tx). Assuming X is continuous.

B) If X~N(0,1), then Mx(t)=e^((t^2)/2) for all t.
Thus, by A), P(X>=x)<=e^((t^2)/2 - tx).
For x=2, find the value of t for which this inequality is sharpest.

2. I'm assuming $M_x(t)$ is the moment generating function. You might want to write out the expression for the moment generating function explicitly in integral form. Similarly, express $P(X\geq x)$ as an integral. Compare the two and play around with it to get the desired inequality.

3. Originally Posted by ninano1205
A) Prove Bernstein's inequality: If X is a random variable for which Mx(t) exists for some t>0, the P(X>=x) <= Mx(t)e^(-tx). Assuming X is continuous.

B) If X~N(0,1), then Mx(t)=e^((t^2)/2) for all t.
Thus, by A), P(X>=x)<=e^((t^2)/2 - tx).
For x=2, find the value of t for which this inequality is sharpest.
For A), you can write, for $t>0$, $P(X\geq x)=P(e^{tX}\geq e^{tx})$ and apply Markov inequality.
In B), you are asked for the $t$ for which the given right-hand side is minimum, so you have to study the variations of the function $t\mapsto e^{(t^2)/2 - tx}$ (differentiate, study the sign of the derivative,...) to find which value of $t$ minimizes it.