• Aug 6th 2008, 11:36 AM
flower3
depending on the moment generating function of the random variable X : M(t)=e^(t^2 ) :find

1. p(x=2)
2.find the mean of X
3. FIND THE VARIANCE of -2x
4.find the the second moment about the origin

i don't know how i starting ??(Headbang)
• Aug 6th 2008, 08:15 PM
mr fantastic
Quote:

Originally Posted by flower3
depending on the moment generating function of the random variable X : M(t)=e^(t^2 ) :find

1. p(x=2)
2.find the mean of X
3. FIND THE VARIANCE of -2x
4.find the the second moment about the origin

i don't know how i starting ??(Headbang)

You're expected to know the following formula: $E\left( X^n \right) = \left. \frac{d^n M}{d t^n}\right| _{t=0}$

2. $\mu = E(X) = \left. \frac{d M}{d t}\right| _{t=0} = \, ....$

4. $E \left( X^2\right) = \left. \frac{d^2 M}{d t^2}\right| _{t=0} = \, ....$

3. $Var(-2X) = (-2)^2 Var(X) = 4 Var(X)$ and $Var(X) = E \left( X^2\right) - \mu^2$.

Now substitute the results from 2. and 4.

1. If X is a continuous random variable then Pr(X = a) = 0.

I have more to say and when I have the time I'll say it.
• Aug 6th 2008, 08:51 PM
mr fantastic
Quote:

Originally Posted by flower3
depending on the moment generating function of the random variable X : M(t)=e^(t^2 )

[snip]

Note that the normal distribution with mean $\mu$ and variance $\sigma^2$ has the moment generating function $e^{\mu t + \frac{t^2 \sigma^2}{2}}$.

Compare this with the given moment generating function and you see that X must follow a normal distribution with $\mu = 0$ and variance $\sigma^2 = 2$.

This gives you a way of checking the answers to 2. and 3.

It also lets you calculate Pr(a < X < b) were it to be asked.
• Aug 7th 2008, 04:51 AM
flower3
(Heart)thank you i think you're the best "mr fantastic"(Clapping)
• Aug 7th 2008, 05:00 AM
flower3
wait >>>>>>

you say the distribution is continuous :

when i find the variance by the 1st and the 2nd derivative for moment generating function i get var(x)=0 but you say that is = 2 how???(Crying)(Crying)
• Aug 7th 2008, 05:09 AM
mr fantastic
Quote:

Originally Posted by flower3
wait >>>>>>

you say the distribution is continuous :

when i find the variance by the 1st and the 2nd derivative for moment generating function i get var(x)=0 but you say that is = 2 how???(Crying)(Crying)

If you showed your working I could identify the mistake(s) you made in getting Var(X) = 0.

$\frac{d^2 M}{dt^2} = 2 e^{t^2} + 4t^2 e^{t^2}$. Substitute t = 0 and you get $E\left(X^2\right) = \left.\frac{d^2 M}{dt^2} \right|_{t=0} = 2$.

Therefore $\sigma^2 = Var(X) = E(X^2) - \mu^2 = 2 - 0 = 2$, as expected.

Therefore Var(-2X) = 4 Var(X) = 8.

Feel free to repeat post #4 now lol!
• Aug 7th 2008, 07:13 AM
flower3
i'm so sorry i solve it before i open my mail and see the solution >>>

i forgot that the derivative for 2te^(t^2 ) is equal to "derivative for two multiply function " so the answer is false >>>>>

but i solve it >>>>>then i verify from your saying

thank you>>>>you're a nice man(Heart)
• Aug 7th 2008, 07:16 AM
flower3
Quote:

Originally Posted by mr fantastic

Therefore Var(-2X) = 2 Var(X) = 4.

this answer is false the truth is :

Therefore Var(-2X) = 4 Var(X) = 8.
• Aug 7th 2008, 04:28 PM
mr fantastic
Quote:

Originally Posted by flower3
this answer is false the truth is :

Therefore Var(-2X) = 4 Var(X) = 8.

Whoops. My mistake. I was thinking sd while typing it. Congratulations for realising the mistake instead of getting all worried and antsy.
• Aug 8th 2008, 03:09 AM
flower3
thank you !(Talking)

all people maybe miscalculate since we're human(Itwasntme)