| Line 20: | Line 20: | ||
---- | ---- | ||
===Answer 1=== | ===Answer 1=== | ||
| + | Firstly, to find the constant k, we can integrate the PDF from negative infinity to positive infinity. The result should equal 1. | ||
| + | |||
| + | <math> | ||
| + | \int_{-\infty}^{\infty} \! f_X(x) \, \mathrm{d}x = 1. | ||
| + | </math> | ||
| + | |||
| + | However, since the limits of the function don't extend all the way out to infinity, we can simply use the lower limit of a and upper limit of b. | ||
| + | |||
| + | <math> | ||
| + | => \int_{a}^{b} \! k \, \mathrm{d}x = 1. | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | => x*k|_a^b = 1 | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | => k = \frac{1}{(b-a)} | ||
| + | </math> | ||
| + | |||
The mean of a random variable is defined as: | The mean of a random variable is defined as: | ||
| Line 37: | Line 57: | ||
= 1/2x^2k|_a^b | = 1/2x^2k|_a^b | ||
</math> | </math> | ||
| − | |||
<math> | <math> | ||
= \frac{k}{2}(b^2-a^2) | = \frac{k}{2}(b^2-a^2) | ||
</math> | </math> | ||
| + | |||
| + | Putting in our value for k, we get: | ||
| + | |||
| + | <math> | ||
| + | = \frac{1}{2(b-a)}(b^2-a^2) | ||
| + | </math> | ||
| + | |||
===Answer 2=== | ===Answer 2=== | ||
Latest revision as of 15:49, 22 March 2013
Contents
Practice Problem: normalizing the probability mass function of a continuous random variable
A random variable X has the following probability density function:
$ f_X (x) = \left\{ \begin{array}{ll} k, & \text{ if } a\leq x \leq b,\\ 0, & \text{ else}, \end{array} \right. $
where k is a constant. Compute the mean of X.
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
Firstly, to find the constant k, we can integrate the PDF from negative infinity to positive infinity. The result should equal 1.
$ \int_{-\infty}^{\infty} \! f_X(x) \, \mathrm{d}x = 1. $
However, since the limits of the function don't extend all the way out to infinity, we can simply use the lower limit of a and upper limit of b.
$ => \int_{a}^{b} \! k \, \mathrm{d}x = 1. $
$ => x*k|_a^b = 1 $
$ => k = \frac{1}{(b-a)} $
The mean of a random variable is defined as:
$ \int \! x*f_X(x) \, \mathrm{d}x. $
Since the probability density function is k on the interval a to b and zero everywhere else, we can simply write:
$ => \int_a^b \! x*k \, \mathrm{d}x. $
Thus solving we get
$ = 1/2x^2k|_a^b $
$ = \frac{k}{2}(b^2-a^2) $
Putting in our value for k, we get:
$ = \frac{1}{2(b-a)}(b^2-a^2) $
Answer 2
Write it here
Answer 3
Write it here.
