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===Answer 1=== | ===Answer 1=== | ||
| − | + | To find the CDF given the PDF, we must integrate: | |
| + | |||
| + | <math>F_X(x) = \int_{-\infty}^x \! f(t) \, \mathrm{d}t.</math> | ||
| + | |||
| + | However, before we integrate, we can setup our solution by knowing some properties of a cumulative distribution. | ||
| + | Since we know that the cumulative distribution varies from 0 to 1 and that the provided pdf has a probability only in the range from a to b, | ||
| + | we can infer that for any x less than a, the CDF will equal 0 and for any x greater than or equal to b, the CDF will equal 1 giving us this solution so far: | ||
| + | |||
| + | <math> F_X (x) = \left\{ | ||
| + | \begin{array}{ll} | ||
| + | 0, & \text{ if } x < a,\\ | ||
| + | ?, & \text{ if } a\leq x < b,\\ | ||
| + | 1, & \text{ if } x \geq b, | ||
| + | \end{array} | ||
| + | \right. </math> | ||
| + | |||
| + | To solve for the between a and b, we perform the integral, however we do not need to integrate from negative infinity, we can simply integrate from the lower limit of a: | ||
| + | |||
| + | <math>\int_{a}^x \! k \, \mathrm{d}t.</math> | ||
| + | |||
| + | Computing the integral we obtain: | ||
| + | |||
| + | <math>kt\vert_a^x = k(x-a)</math> | ||
| + | |||
| + | Thus, the CDF is: | ||
| + | |||
| + | <math> F_X (x) = \left\{ | ||
| + | \begin{array}{ll} | ||
| + | 0, & \text{ if } x < a,\\ | ||
| + | k(x-a), & \text{ if } a\leq x < b,\\ | ||
| + | 1, & \text{ if } x \geq b, | ||
| + | \end{array} | ||
| + | \right. </math> | ||
| + | |||
===Answer 2=== | ===Answer 2=== | ||
Write it here | Write it here | ||
Revision as of 06:39, 2 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. Determine the cumulative distribution function 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
To find the CDF given the PDF, we must integrate:
$ F_X(x) = \int_{-\infty}^x \! f(t) \, \mathrm{d}t. $
However, before we integrate, we can setup our solution by knowing some properties of a cumulative distribution. Since we know that the cumulative distribution varies from 0 to 1 and that the provided pdf has a probability only in the range from a to b, we can infer that for any x less than a, the CDF will equal 0 and for any x greater than or equal to b, the CDF will equal 1 giving us this solution so far:
$ F_X (x) = \left\{ \begin{array}{ll} 0, & \text{ if } x < a,\\ ?, & \text{ if } a\leq x < b,\\ 1, & \text{ if } x \geq b, \end{array} \right. $
To solve for the between a and b, we perform the integral, however we do not need to integrate from negative infinity, we can simply integrate from the lower limit of a:
$ \int_{a}^x \! k \, \mathrm{d}t. $
Computing the integral we obtain:
$ kt\vert_a^x = k(x-a) $
Thus, the CDF is:
$ F_X (x) = \left\{ \begin{array}{ll} 0, & \text{ if } x < a,\\ k(x-a), & \text{ if } a\leq x < b,\\ 1, & \text{ if } x \geq b, \end{array} \right. $
Answer 2
Write it here
Answer 3
Write it here.
