# Thread: Distribution to density to expected

1. ## Distribution to density to expected

A random variable X has the cumulative distribution function
$\displaystyle F(x) = 0$ for $\displaystyle x<1$
$\displaystyle F(x) = \frac{x^2 - 2x + 2}{2}$ for $\displaystyle 1 \leq x < 2$
$\displaystyle F(x) = 1$ for $\displaystyle x \geq 2$

Calculate the variance of X.

The answer specifies that the density function is
$\displaystyle f(x) = 0.5$ if x=1
$\displaystyle f(x) = x-1$ if 1<x<2
$\displaystyle f(x) = 0$ otherwise

Then
$\displaystyle E(X) = 0.5 + \int_1^2 x(x-1) dx$
$\displaystyle E(X^2) = 0.5 + \int_1^2 x^2(x-1) dx$

I got the f(x) = x-1 part, and I got how to calculate the variance after you have the expected values, but I'm lost on other questions.

My questions are:
Where do we get $\displaystyle f(x) = 0.5$ if x=1? F(1) = 0.5, but I can't figure out why f(1) would equal 0.5.

What is the rule for putting parts of the stepwise density function into the expected value equations? I don't know what the rule is called so I don't know how to review it. We're adding the slope at a single point to the slope over a big area, which is something I can't quite work out visually.

2. Originally Posted by Boris B
A random variable X has the cumulative distribution function
$\displaystyle F(x) = 0$ for $\displaystyle x<1$
$\displaystyle F(x) = \frac{x^2 - 2x + 2}{2}$ for $\displaystyle 1 \leq x < 2$
$\displaystyle F(x) = 1$ for $\displaystyle x \geq 2$

Calculate the variance of X.

The answer specifies that the density fucntion is
$\displaystyle f(x) = 0.5$ if x=1
$\displaystyle f(x) = x-1$ if 1<x<2
$\displaystyle f(x) = 0$ otherwise

Then
$\displaystyle E(X) = 0.5 + \int_1^2 x(x-1) dx$
$\displaystyle E(X^2) = 0.5 + \int_1^2 x^2(x-1) dx$

I got the f(x) = x-1 part, and I got how to calculate the variance after you have the expected values, but I'm lost on other questions.

My questions are:
Where do we get $\displaystyle f(x) = 0.5$ if x=1? F(1) = 0.5, but I can't figure out why f(1) would equal 0.5.

What is the rule for putting parts of the stepwise density function into the expected value equations? I don't know what the rule is called so I don't know how to review it. We're adding the slope at a single point to the slope over a big area, which is something I can't quite work out visually.
$\displaystyle F(x) = 0$ for $\displaystyle x<1$ implies that $\displaystyle \Pr(X < 1) = 0$.

$\displaystyle F(x) = \frac{x^2 - 2x + 2}{2}$ for $\displaystyle 1 \leq x < 2$ implies that $\displaystyle \Pr(X \leq 1) = \frac{1^2 - 2(1) + 2}{2} = \frac{1}{2}$.

It follows that $\displaystyle \Pr(X = 1) = \Pr(X \leq 1) - \Pr(X < 1) = \frac{1}{2}$.

3. Originally Posted by Boris B
A random variable X has the cumulative distribution function
$\displaystyle F(x) = 0$ for $\displaystyle x<1$
$\displaystyle F(x) = \frac{x^2 - 2x + 2}{2}$ for $\displaystyle 1 \leq x < 2$
$\displaystyle F(x) = 1$ for $\displaystyle x \geq 2$

Calculate the variance of X.

The answer specifies that the density fucntion is
$\displaystyle f(x) = 0.5$ if x=1
$\displaystyle f(x) = x-1$ if 1<x<2
$\displaystyle f(x) = 0$ otherwise

Then
$\displaystyle E(X) = 0.5 + \int_1^2 x(x-1) dx$
$\displaystyle E(X^2) = 0.5 + \int_1^2 x^2(x-1) dx$

I got the f(x) = x-1 part, and I got how to calculate the variance after you have the expected values, but I'm lost on other questions.

My questions are:
Where do we get $\displaystyle f(x) = 0.5$ if x=1? F(1) = 0.5, but I can't figure out why f(1) would equal 0.5.
The $\displaystyle 1/2$ comes from the jump discontinuity in the cumulative distribution $\displaystyle F(x)$ at $\displaystyle x=1$, indicating a probability mass of $\displaystyle 1/2$ at that point.

Another way of thinking about this is to think of the density as a generalised function, then we may represent the density of a piecewise continuous cumulative distribution as the sum of a continuous function and delta functionals at the discontiuities of amplitude equal to the size of the jumps.

RonL

4. Originally Posted by Boris B
A random variable X has the cumulative distribution function
$\displaystyle F(x) = 0$ for $\displaystyle x<1$
$\displaystyle F(x) = \frac{x^2 - 2x + 2}{2}$ for $\displaystyle 1 \leq x < 2$
$\displaystyle F(x) = 1$ for $\displaystyle x \geq 2$

Calculate the variance of X.

The answer specifies that the density fucntion is
$\displaystyle f(x) = 0.5$ if x=1
$\displaystyle f(x) = x-1$ if 1<x<2
$\displaystyle f(x) = 0$ otherwise
If this last is the given answer for the "density" then it is wrong, as as given it is not a density since:

$\displaystyle \int_{-\infty}^{\infty} f(x)~dx=1/2$

The density is the generalised function:

$\displaystyle f(x)=g(x)+(1/2)\delta(x-1)$

where

$\displaystyle g(x) = x-1, \ \ 1<x<2.$
$\displaystyle g(x) = 0, \ \ \mbox{otherwise}.$

RonL

5. Originally Posted by Boris B

What is the rule for putting parts of the stepwise density function into the expected value equations? I don't know what the rule is called so I don't know how to review it. We're adding the slope at a single point to the slope over a big area, which is something I can't quite work out visually.
If you use the generalised function approach you have a perfectly normal looking equation for the expected value. The problem arrises if you represent the distribution as mixed discrete/continuous, when you have to work with the two components seperately.

RonL