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#.

1 Answer
Dec 29, 2017

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#.