rayne said:Given the PDF:
f(x) = 1/12 , 0 < x <= 3
x/18, 3 < x <= 6
0, otherwise
find the expected value, E(x).
I know how to find the expected value if there was only one interval, but don't how to do it for two.
chisigma said:The integral can be devided in two integrals as follows...
$\displaystyle E \{X \} = \frac{1}{12}\ \int_{0}^{3} x\ d x + \frac{1}{18}\ \int_{3}^{6} x^{2}\ dx\ (1)$
Kind regards
$\chi$ $\sigma$
The expected value of a continuous random variable is a measure of the average value that can be expected from the variable over a large number of trials or observations. It is a weighted average of all possible values that the variable can take on, with the weighting factor being the probability of each value occurring.
The expected value of a continuous random variable is calculated by multiplying each possible value by its corresponding probability of occurring, and then summing these products. This can also be expressed as the integral of the variable multiplied by its probability density function.
The expected value of a continuous random variable is important because it provides a way to summarize the behavior of the variable and make predictions about its future values. It also allows for comparison between different variables and can help in decision-making processes.
Expected value and variance are both measures of central tendency and variability, respectively, for a continuous random variable. They are related in that the variance is a measure of the spread of the variable's values around its expected value. A higher variance indicates a greater spread, while a lower variance indicates a more concentrated distribution around the expected value.
Yes, the expected value of a continuous random variable can be negative if the distribution of the variable is skewed to the left, meaning that there is a higher probability of lower values occurring. However, in most cases, the expected value is positive as it represents the average value of the variable.