Question

May I know how to solve it?? Thank you

The moment generating function of X is M(t)=`(e^2t-e^6t)//t(6-2)`.
a) what is the distribution of X?
b) find P(2`<=X<=3)`.
c)the p.d.f for uniform distribution is f(x)= 1/(b-a) for `a<=x<=b`.
Show that the `mu=(a+b)//2 and theta^2=(b-a)^2//12`.

Answer 1

Recognize that `M(t)=(e^{6t}-e^{2t})/(4t)` is the moment generating function of a uniform distribution over the interval `[2,6]`. Find the probability by integrating the p.d.f. Calculate `mu=E[X]`, `E[X^{2}]`, and `sigma^{2}=E[X^{2}]-mu^{2}`. (You could also try, with computer help, to calculate `M'(0)` and `M''(0)` (as limits) to help you confirm the mean and variance.)

Explanation:

As you stated, for a continuous random variable `X` uniform distribution over `[a,b]` (for `a <= x <= b`), the p.d.f. is `f(x)=1/(b-a)`.

The moment generating function is

`M(t)=E[e^{tX}]=int_{-infty}^{infty}e^{tx}f(x)dx=1/(b-a)int_{a}^{b}e^{tx}dx`

`=1/((b-a)t)e^{tx}|_{x=a}^{x=b}=(e^{bt}-e^{at})/((b-a)t)`.

(a) Therefore, `M(t)=(e^{6t}-e^{2t})/(4t)` is the moment generating function of a uniform distribution over the interval `[2,6]` (i.e., `f(x)=1/4` for `2 <= x <= 6`).

(b) The probability is `P(2 <=X <=3 ) = int_{2}^{3}f(x)dx=1/4`.

(c) In the general case, the mean is

`mu=E[X]=int_{-infty}^{infty}xf(x)dx=1/(b-a) int_{a}^{b}x dx=1/(2(b-a))x^{2}|_{a}^{b}=(b^{2}-a^{2})/(2(b-a))=(a+b)/2`

The mean is also the "first moment" `mu=E[X]=lim_{t->0}M'(t)` (see https://en.wikipedia.org/wiki/Moment-generating_function). Use a computer-algebra system to help you calculate this if you are interested.

The second moment is `E[X^{2}]` (which also equals `lim_{t->0}M''(t)`). This is

`E[X^{2}]=int_{-infty}^{infty}x^{2}f(x)dx=1/(b-a) int_{a}^{b}x^{2} dx=1/(3(b-a))x^{3}|_{a}^{b}=(b^{3}-a^{3})/(3(b-a))=(a^{2}+ab+b^{2})/3`

The variance `sigma^{2}` (or `theta^{2}`) is

`sigma^{2}=E[X^{2}]-mu^{2}=(a^{2}+ab+b^{2})/3-(a^{2}+2ab+b^{2})/4`

`=(4a^{2}+4ab+4b^{2}-3a^{2}-6ab-3b^{2})/12=(a^{2}-2ab+b^{2})/12`

`=((a-b)^{2})/12=((b-a)^{2})/12`.