Distribution to density to expected

**Boris B** · April 23rd 2008, 08:41 PM

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

Calculate the variance of X.

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

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

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 $f(x) = 0.5$ 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.

**mr fantastic** · April 23rd 2008, 09:02 PM

Originally Posted by Boris B

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

Calculate the variance of X.

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

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

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 $f(x) = 0.5$ 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.

$F(x) = 0$ for $x<1$ implies that $\Pr(X < 1) = 0$ .

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

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

**CaptainBlack** · April 23rd 2008, 10:55 PM

Originally Posted by Boris B

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

Calculate the variance of X.

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

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

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 $f(x) = 0.5$ if x=1? F(1) = 0.5, but I can't figure out why f(1) would equal 0.5.

The $1/2$ comes from the jump discontinuity in the cumulative distribution $F(x)$ at $x=1$ , indicating a probability mass of $1/2$ 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

**CaptainBlack** · April 23rd 2008, 11:04 PM

Originally Posted by Boris B

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

Calculate the variance of X.

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

If this last is the given answer for the "density" then it is wrong, as as given it is not a density since:

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

The density is the generalised function:

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

where

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

RonL

**CaptainBlack** · April 23rd 2008, 11:07 PM

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

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