Question

how to find moment generating function? Anyone can explain with the example that if X denotes the out come when fair die is tossed find MGF, mean and variance.

Answer 1

@hartnn buddy can you help me with this?

Answer 2

mathworld.wolfram.com/Moment-GeneratingFunction.html

Answer 3

@Hero can you help me with this?

Answer 4

@zepdrix can you help me?

Answer 5

@electrokid

Answer 6

@mathstudent55

Answer 7

@amistre64

Answer 8

@jim_thompson5910

Answer 9

@Preetha

Answer 10

anyone? O_________O

Answer 11

https://onlinecourses.science.psu.edu/stat414/node/52

im reading thru this to see if i can learn about MGFs

Answer 12

\[\mu=E(X)=\frac16\]might be needed

Answer 13

\[E(X^2)=\sigma^2+\mu^2\]seems line another

Answer 14

hey @amistre64 check this from same site
https://onlinecourses.science.psu.edu/stat414/node/72

Answer 15

already there :)

Answer 16

p = 1/6, 1-p = 5/6

Answer 17

M(t)=E(etX)=∑ e^tx * f(x)
I am not getting whose summation is to be taken?

Answer 18

im not sure at the moment why the e^(tx) is important

Answer 19

o_o because it is in the formula for mgf= M(t)

Answer 20

still if you have something else please go on..

Answer 21

yeah, i see that :) i just like to know why its there ...

the function is the probability distribution; which in this case is uniform. f(x) = 1/6, from 0 to 6



and zero everywhere else

Answer 22

yep thats correct

Answer 23

or could we go with the binomial of p=1/6, 1-p = 5/6 ?

Answer 24

anything, however you want. I just want to learn you get the MGF here I'll understand the thing.

Answer 25

well, from what i see, we take that e^(tx) and multiply it to our probability distribution function. and simplify

Answer 26

\[f(x)=\frac16~:~(0,6)\]
but thats a continuous random variable setup .... seeing how there is no real way to get x=3.5, this might not be the best setup for f(x)

Answer 27

:( what to do then?

Answer 28

hmmm, if we wanted to know the probability of getting exactly one 3, out of 5 tosses of the die, we would run a binomial setup

ffffs
fffsf
ffsff
fsfff
sffff

these are the number of ways to get exactly 1 "value" out of 5 tosses; s = success, f = fail

5 * (s)^1* (f)^4\[\binom{5}{1}(\frac16)^1(\frac56)^4\]

Answer 29

so it depends on what it is we are trying to determine here

Answer 30

\[\sum_{x=0}^6~ \frac16e^{tx}\]doesnt seem to mean much to me

Answer 31

yup this one I know... and I think we are going correct as per the video
M(t) = ( (1-p) + p* e^t ) ^n
so here
M(t) = ( 5/6 + (1/6)* e^t )^n
must be the MGF

Answer 32

i agree

Answer 33

then I think 1st derivative of mgf with t=0 give mean

Answer 34

dont know what to do. am I right???

Answer 35

derivatives with respect to "t", yes ... evaluated at t=0

Answer 36

what do you get?? I got 1 as mean o.O. is is right?

Answer 37

assuming we want n=1\[1[\frac56+\frac16e^{0}]^{0}~\frac16e^0\]
\[1(1)\frac16(1)=\frac16\]

Answer 38

ahan thats correct general formula for mean is n*p here n is 1 and p is 1/6 so mean will be 1/6. Thats absolutely correct. :)

Answer 39

now can you find variance in same way ?

Answer 40

variance is the second derivative w.r.t.t
\[M''(t)=n(n-1)[(1-p)+pe^{t}]^{n-2}~p^2e^{2t}+n[(1-p)+pe^{t}]^{n-1}~pe^{t}\]
\[M''(0)=n(n-1)[(1-p)+p]^{n-2}~p^2+n[(1-p)+p]^{n-1}~p\]

Answer 41

I think variance should be 2nd derivative minus mean square so in the above equation as we have n=1 1st term will be zero and 2nd term 1/6 and variance will be
1/6- (1/6)^2 = 5/36

Answer 42

correct me if I am wrong

Answer 43

variance = np(1-p) = 5/36 .. yes, so i have to consider it i did the derivative correctly

Answer 44

:D yup thats correct. we have done it. Thanks buddy.

Answer 45

youre welcome