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.

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dfbustos dfbustos · Answer 1 · 2019-12-24T18:50:08+01:00

Answer:

a) [tex]E(X)=\frac{b^2 -a^2}{2(b-a)}=\frac{(b-a)(b+a)}{2*(b-a)}=\frac{b+a}{2}[/tex]

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

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

[tex]E(X^2)=\int_{a}^b \frac{x^2}{b-a}dx = \frac{1}{3(b-a)} x^3 \Big|_a^b \[/tex]

[tex]E(X) = \frac{b^3 -a^3}{3(b-a)}=\frac{(b-a)(b^2 +ab +a^2)}{3*(b-a)}=\frac{b^2+ab+a^2}{3}[/tex]

And now we can find the variance like this:

[tex]Var(X) =\frac{b^2+ab+a^2}{3} -[\frac{b+a}{2}]^2[/tex]

[tex]Var(X)=\frac{4b^2 +4ab +4a^2 -3b^2-3a^2-6ab}{12}=\frac{b^2 -2ab +a^2}{12}=\frac{(b-a)^2}{12}[/tex]

c) [tex]F(X)=\int_a^x \frac{1}{b-a} dr =\frac{r}{b-a} \Big|_a^x \ =\frac{x-a}{b-a}[/tex]

Step-by-step explanation:

For this case we assume that [tex]X \sim Unif(a,b)[/tex]

And the density function is given by [tex]f(X)=\frac{1}{b-a}[/tex]

Part a

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

[tex]E(X)=\int_{a}^b \frac{x}{b-a}dx = \frac{1}{2(b-a)} x^2 \Big|_a^b \[/tex]

[tex]E(X) = \frac{b^2 -a^2}{2(b-a)}=\frac{(b-a)(b+a)}{2*(b-a)}=\frac{b+a}{2}[/tex]

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:

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

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

[tex]E(X^2)=\int_{a}^b \frac{x^2}{b-a}dx = \frac{1}{3(b-a)} x^3 \Big|_a^b \[/tex]

[tex]E(X) = \frac{b^3 -a^3}{3(b-a)}=\frac{(b-a)(b^2 +ab +a^2)}{3*(b-a)}=\frac{b^2+ab+a^2}{3}[/tex]

And now we can find the variance like this:

[tex]Var(X) =\frac{b^2+ab+a^2}{3} -[\frac{b+a}{2}]^2[/tex]

[tex]Var(X)=\frac{4b^2 +4ab +4a^2 -3b^2-3a^2-6ab}{12}=\frac{b^2 -2ab +a^2}{12}=\frac{(b-a)^2}{12}[/tex]

Part c

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

[tex]F(X)=\int_a^x \frac{1}{b-a} dr =\frac{r}{b-a} \Big|_a^x \ =\frac{x-a}{b-a}[/tex]