Thank you for posing this question. It is an interesting question with
a surprisingly simple and elegant solution.
The function f(t) you define is known as the moment generating
function for a normal
distribution. http://www.plmsc.psu.edu/~www/matsc597/probability/moments/moments.html
.
Once you compute the moment generating function, it turns out you can
easily compute all the moments of a distribution.
The actual computation of the moment generating function for a normal
distribution is performed in the course notes:
http://www.qmw.ac.uk/~ugte133/courses/MATHSTAT/6normgf.pdf . (You will
need a pdf reader to view this URL, which you can get from this URL:
http://www.adobe.com/products/acrobat/readstep.html .
Please see equations 2, 3, and 4 on page 1 of that document.
If you have any questions about their derivation, I would be happy to
answer them. I am not redoing their work here since actually integrals
and quotients can be a little bit hard to read in text.
The answer turns out to be, by the way,
f(t)= e^( (t^2)/2)
To use this value to compute the expectation of Z^4, we follow the
derivation on the presentation of moment generating functions at:
http://www.plmsc.psu.edu/~www/matsc597/probability/moments/moments.html
and specifically: http://www.plmsc.psu.edu/~www/matsc597/probability/moments/node4.html
.
Recall that
(*) e^x= 1 + x + x^2/2! + x^3 /3! ...
Hence,
f(t) = E(e^(tZ))
= E( 1 + tZ + t^2/2! Z^2 + ...)
= E(1) + E(tZ) + E(t^2/2! Z^2 + ...)
(**) = 1 + tE(Z) + t^2/2! E(Z^2) + t^3/3! E(Z^3) + t^4/4! E(Z^4) +
....
But we know already that
f(t)= e^( (t^2)/2)
= 1 + t^2/2 + (t^2/2)^2/2! + ...
= 1 + t^2/2 + t^4/8 + ...
from the defintion of exponetial (*) above.
We now have two expressions for f(t), the above and that of (**) .
Equating the coefficients of t^4, we see that
E(Z^4) /4! = 1/8
so
E(Z^4) = 4!/8 = 3 .
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Request for Answer Clarification by
brkyrhrt99_00ga
on
05 Oct 2002 08:25 PDT
I am rerally unclear at how you derive at 3. I understand how you
find the moment generating function. But from then on I am lost

Clarification of Answer by
rbnnga
on
05 Oct 2002 10:15 PDT
I'd be happy to clarify, but first, I don't know what you mean by the
comment:
"I am rerally unclear at how you derive at 3."
What is 3?
Also, do you mean that you are ok with the derivation that
f(t)=E(e^(t^2)/2) ?
If so, just tell me where things start to get unclear, i.e. which
line.
(By the way when I first saw this proof it also seemed somehow magical
and strange to me; all this machinery that doesn't have anything to do
with E(Z^4) somehow E(Z^4) just falls out.

Request for Answer Clarification by
brkyrhrt99_00ga
on
05 Oct 2002 10:52 PDT
this is where I am lost:
(*) e^x= 1 + x + x^2/2! + x^3 /3! ...
Hence,
f(t) = E(e^(tZ))
= E( 1 + tZ + t^2/2! Z^2 + ...)
= E(1) + E(tZ) + E(t^2/2! Z^2 + ...)
(**) = 1 + tE(Z) + t^2/2! E(Z^2) + t^3/3! E(Z^3) + t^4/4! E(Z^4) +
....
But we know already that
f(t)= e^( (t^2)/2)
= 1 + t^2/2 + (t^2/2)^2/2! + ...
= 1 + t^2/2 + t^4/8 + ...
from the defintion of exponetial (*) above.
We now have two expressions for f(t), the above and that of (**) .
Equating the coefficients of t^4, we see that
E(Z^4) /4! = 1/8
so
E(Z^4) = 4!/8 = 3 .

Clarification of Answer by
rbnnga
on
05 Oct 2002 13:44 PDT
I will begin by recalling some basic facts about Taylor series.
[Taylor Series fact] Every infinitely differentiable function g(x) can
be written in a Taylor series:
g(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ...
where
a_i is constant value equal to the i'th derivative of g, evaluated at
0, divided by i! (i!, of course, is 1 * 2 * 3 *...* i ):
a_0 = g(0)
a_1 = g'(0)/1!
a_2 = g''(0)/2!
a_3 = g'''(0)/3!
...
[Power series fact] If it is also the case that
g(x) = b_0 + b_1 x + b_2 x^2 + b_3 x^3 + ...
then a_i = b_i everywhere .
The above are some facts from calculus that we will use in the
following.
I will now clarify my original derivation, with new comments in
brackets.
(*) e^x= 1 + x + x^2/2! + x^3 /3! ...
[This is the taylor series expansion of e^x . It holds because the
i'th derivative of e^x equals e^x; when evaluated at 0, therefore, the
i'th derivative of e^x is therefore equal to e^0 which is 1. It
follows from the formula after "Taylor series fact" above]
Hence,
f(t) = E(e^(tZ))
[This is the definition of f(t) ]
= E( 1 + tZ + t^2/2! Z^2 + ...)
[We are simply rewriting e^(tZ) using the formula (*), where "x" is
replaced by "tZ"]
= E(1) + E(tZ) + E(t^2/2! Z^2) + ...
[We are using the fact that the sum of the expectations of random
variables is equal to the expectation of the sum of these variables]
(**) = 1 + tE(Z) + t^2/2! E(Z^2) + t^3/3! E(Z^3) + t^4/4! E(Z^4) +
....
[Here, remember that t is a a constant. So we are using that the fact
that, if Y is a random variable, and if a is any constant, then E( aY
) = a E(Y) .]
But we know already that
f(t)= e^( (t^2)/2)
[We now have two expressions for f(t). The one in ** and the one we
proved, this one, in the first part of the problem ]
= 1 + t^2/2 + (t^2/2)^2/2! + ...
[This uses the expression (*) above, with x replaced by t^2/2 .]
[Let's call the following equation (***)]
= 1 + t^2/2 + t^4/8 + ...
[This just writes out the first few terms. We are only interested in
the coefficient of t^4.]
from the definition of exponetial (*) above.
We now have two expressions for f(t), the above and that of (**) .
[The first expression for f(t) that we found is expression (**) and
the second is expression (***):
(**) f(t)= 1 + tE(Z) + t^2/2! E(Z^2) + t^3/3! E(Z^3) + t^4/4! E(Z^4) +
....
(**) f(t)= 1 + 0t +t^2/2 + 0t^3 + t^4/8 +
...
]
Equating the coefficients of t^4, we see that
[Recall from "power series fact" above that if two power series equal
the same funtion, the coefficients of each term must be equal. Hence,
the coefficients of t^4 in the two expressions for f(t) above must be
equal. These coefficients are E(Z^4)/4! and 1/8 for (**) and (***)
respectively. Hence: ]
E(Z^4) /4! = 1/8
so
E(Z^4) = 4!/8 = 3 .
[We just multiplied
[That is since 4!/8 = 4 * 3 * 2 * 1 / 8 = 3]
