Calculus...?

Question

The density function of X is given by

f(x) = a + bx^2 if 0 ≤ x ≤ 1
f(x) = 0 otherwise

If the expectation of X is E(X) = 5, find a and b.
a = _______
b = _______

Answer 1

Integrate both the density and the expectation functions over the interval [0, 1].

∫f(x)dx = ∫(a + bx²)dx = ax + (1/3)bx³ | [Eval from 0 to 1]

= a + b/3 = 1

Multiply thru by 3 to clear the denominators.

3a + b = 3
________

∫xf(x)dx = ∫x(a + bx²)dx = ∫(ax + bx³)dx

= [(1/2)ax² + (1/4)bx^4] | [Eval from 0 to 1]

= a/2 + b/4 = 5

Multiply thru by 4 to clear the denominators.

2a + b = 20
_________

Now we have two equations and two unknowns.

3a + b = 3
2a + b = 20

Subtract the second equation from the first.

a = -17

Plug the value for a into one of the equations and solve for b.

2a + b = 20
b = 20 - 2a = 20 - 2(-17) = 20 + 34 = 54

So a = -17 and b = 54.

Answer 2

Trevor, this isn't possible at all.

The probability density function means to indicate the relative probability that x lies in the region shown. For example, the probability that x would have any value outside the range [0,1], is given as zero. This means that x can simply not have a value outside this range.

How then, can the average or expected value of x turn out to be 5?

I believe there is a typo here. It should be something else.

Let me assume that the expected value is c and solve the problem first. I'll then show you what values can c have. To see the equations below, copy and paste all lines starting from \[ and ending in \] in the online form at http://www.tlhiv.org/cgi-bin/LaTeXpreviewer/index.cgi

Hit the preview button to see the equations clearly.

\[
\int_{0}^{1} {(a+bx^2) dx} = 1
\]
\[
\int_{0}^{1} {(ax+bx^3) dx} = c
\]
\[
{\left \left( ax+\frac{bx^3}{3}\right) \right|}_{0}^{1} = 1 = a + \frac{b}{3}
\]
\[
{\left \left( \frac{ax^2}{2}+\frac{bx^4}{4}\right) \right|}_{0}^{1} = c = \frac{a}{2} + \frac{b}{4}
\]

If you solve the last two equations, we get
\[
a = 3-2c ; b= 6c-6
\]

Now, for this probability density function to be valid, we must have the coefficient of x^2 as negative, so that we have only non-negative probabilities (shaped like an inverted U, with its vertex as a maximum). Also, we must make sure that at both the ends of the range [0,1], the probability density function should be non-negative.

This means:

\[
b < 0
\]
and
\[
a \geq 0
\]
or
\[
a+b \geq 0
\]

This would therefore mean that:

\[
c \in \left[0.75,1\right)
\]
or
\[
c \in \left(0,\frac{2}{3}\right]
\]

So, please check your question.

Answer 3

We know that if you integrate the density function over the interval, we get 1. So:

Int (0 to 1) (a + bx^2)dx = 1
ax + (b/3)(x^3) evaluated at (0 to 1) = 1
a + b/3 = 1.

We also know that the expectation is the integral of x*f(x) over the interval. So:
Int (0 to 1) x(a + bx^2)dx = 5
Int (0 to 1) ax + bx^3 dx = 5
(a/2)x^2 + (b/3)x^3 evaluated at (0 to 1) = 5
(a/2) + (b/3) = 5

So you have a system of equations, 2 unknowns:
a + b/3 = 1
a/2 + b/3 = 5

Subtract eq. 2 from eq. 1:
a/2 = -4
a = -8
From eq 1:
-8 + b/3 = 1
b = 27

Answer 4

do your own homework