As a consequence,
possesses a
mgf:
The moment generating function takes its name by the fact that it can be used
to derive the moments of
,
as stated in the following proposition. g. Thus the denominator drops out of the equation. Make that substitution:Cancel out the terms and we have our nice-looking moment-generating function:If we take the derivative of this function and evaluate at 0 we get the mean of the gamma distribution:Recall that is the mean time between events and is the number of events.
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As an example, consider
X
Chi-Squared
{\displaystyle X\sim {\text{Chi-Squared}}}
with
k
{\displaystyle k}
degrees of freedom.
Example
In the previous example we have demonstrated that the mgf of an exponential
random variable
isThe
expected my latest blog post of
can be computed by taking the first derivative of the
see here now evaluating it at
:The
second moment of
can be computed by taking the second derivative of the
mgf:and
evaluating it at
:And
so on for higher moments. Not what you would expect when you start with this:How do we get there? First lets combine the two exponential terms and move the gamma fraction out of the integral:Multiply in the exponential by , add the two terms together and factor out :Now were ready to do so substitution in the integral. If we take the second derivative of the moment-generating function and evaluate at 0, we get the second moment about the origin which we can use to find the variance:Now find the variance:Going back to our example with (number of events) and (mean time between events), we have as our variance .
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” This is because in some cases, the moments exist and yet the moment-generating function does not, because the limit
may not exist. Suppose that \(Y\)has the following mgf. The log-normal distribution is an example of when this occurs. That is, \(M(t)\) is the moment generating function (“m.
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If random variable
X
{\displaystyle X}
has moment generating function
M
X
(
t
)
{\displaystyle M_{X}(t)}
, then
X
+
{\displaystyle \alpha X+\beta }
has moment generating function
M
X
+
(
t
)
=
e
t
M
X
(
t
)
{\displaystyle M_{\alpha X+\beta }(t)=e^{\beta t}M_{X}(\alpha t)}
If
S
n
=
i
=
1
n
a
i
X
i
{\displaystyle S_{n}=\sum _{i=1}^{n}a_{i}X_{i}}
, where the Xi are independent random variables and the ai are constants, then the probability density function for Sn is the convolution of the probability density functions of each of the Xi, and the moment-generating function for Sn is given by
For vector-valued random variables
X
{\displaystyle \mathbf {X} }
with real components, the moment-generating function is given by
where
t
{\displaystyle \mathbf {t} }
is a vector and
,
{\displaystyle \langle \cdot ,\cdot \rangle }
is the dot product. .