By the way, I took this example from the textbook Probability and Statistical Inference (7th ed) by Hogg and Tanis (p. We found that the MGF of the Exponential is \(\frac{\lambda}{\lambda – t}\).
Picking
t
=
m
/
(
2
m
+
k
)
{\displaystyle t=m/(2m+k)}
and plugging into the bound, we get
We know that in this case the correct bound is
E
[
X
m
]
2
m
(
m
+
k
/
2
explanation )
/
(
k
/
2
)
{\displaystyle E[X^{m}]\leq 2^{m}\Gamma (m+k/2)/\Gamma (k/2)}
.
Proposition
Let
and
Go Here be two random variables.
Getting Smart With: Multivariate Statistics
As its name implies, the moment generating function can be used to compute a distribution’s moments: the nth moment about 0 is the nth derivative of the moment-generating function, evaluated at 0. d. If \(X\) is the number of success out of \(n\) trials, then a good estimate of \(p=P(\text{success})\) would be the number of successes out of the total number of trials. .