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Let a < b be real numbers. Let X be a random variable such that it is known that a <= X <= b and for every subinterval of [a; b], the probability that X will belong to that subinterval is proportional to the length of that subinterval. Then we say X has the uniform distribution on the interval [a; b] and the probability density function of X is given by: {1/b-a if a<=x<=b 0 otherwiseand we write X- U(a,b)(a) Find E[X]. Does this agree with your intuition? (b) Show Var[X]= b^2-a^2/12 Hint: b^3- a^3 = (b- a) (b2^+ ab + a^2)(c) Find the c.d.f of X.

1 Answer

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Answer:

a)
E(X)=(b^2 -a^2)/(2(b-a))=((b-a)(b+a))/(2*(b-a))=(b+a)/(2)

b)
Var(X) =E(X^2) -[E(X)]^2

So first we need to find the second central moment like this:


E(X^2)=\int_(a)^b (x^2)/(b-a)dx = (1)/(3(b-a)) x^3 \Big|_a^b \


E(X) = (b^3 -a^3)/(3(b-a))=((b-a)(b^2 +ab +a^2))/(3*(b-a))=(b^2+ab+a^2)/(3)

And now we can find the variance like this:


Var(X) =(b^2+ab+a^2)/(3) -[(b+a)/(2)]^2


Var(X)=(4b^2 +4ab +4a^2 -3b^2-3a^2-6ab)/(12)=(b^2 -2ab +a^2)/(12)=((b-a)^2)/(12)

c)
F(X)=\int_a^x (1)/(b-a) dr =(r)/(b-a) \Big|_a^x \ =(x-a)/(b-a)

Explanation:

For this case we assume that
X \sim Unif(a,b)

And the density function is given by
f(X)=(1)/(b-a)

Part a

In order to find the expected value we need to do the following integral:


E(X)=\int_(a)^b (x)/(b-a)dx = (1)/(2(b-a)) x^2 \Big|_a^b \


E(X) = (b^2 -a^2)/(2(b-a))=((b-a)(b+a))/(2*(b-a))=(b+a)/(2)

And yes agree with the intuition since is the middle value of the interval (a,b)

Part b

For this case we need to use the definition of variance:


Var(X) =E(X^2) -[E(X)]^2

So first we need to find the second central moment like this:


E(X^2)=\int_(a)^b (x^2)/(b-a)dx = (1)/(3(b-a)) x^3 \Big|_a^b \


E(X) = (b^3 -a^3)/(3(b-a))=((b-a)(b^2 +ab +a^2))/(3*(b-a))=(b^2+ab+a^2)/(3)

And now we can find the variance like this:


Var(X) =(b^2+ab+a^2)/(3) -[(b+a)/(2)]^2


Var(X)=(4b^2 +4ab +4a^2 -3b^2-3a^2-6ab)/(12)=(b^2 -2ab +a^2)/(12)=((b-a)^2)/(12)

Part c

In order to find the cumulative distribution function we just ned to do the following integral:


F(X)=\int_a^x (1)/(b-a) dr =(r)/(b-a) \Big|_a^x \ =(x-a)/(b-a)

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