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.