If X is a random variable with the

Frans Vigo

Frans Vigo

Answered question

2022-07-07

If X is a random variable with the following probability density function: (π‘₯) = { 2 |π‘₯| 1 0 Determine: a. c value b. E(X) c. Var(X) d. P( X > )

Answer & Explanation

xleb123

xleb123

Skilled2023-06-02Added 181 answers

To solve the problem, we need to find several properties of the random variable X, given its probability density function (PDF).
a. To determine the c value, we need to integrate the PDF over its entire range and set it equal to 1, as the total area under the PDF must be equal to 1. The given PDF is:
f(x)={2|x|for βˆ’1≀x≀10otherwise
Integrating the PDF from -1 to 1:
βˆ«βˆ’112|x|dx
Splitting the integral into two parts based on the absolute value:
βˆ«βˆ’112xdx+βˆ«βˆ’11βˆ’2xdx
Integrating each part:
[x2]βˆ’11+[βˆ’x2]βˆ’11
Evaluating at the limits:
(12βˆ’(βˆ’1)2)+(βˆ’12βˆ’(βˆ’(βˆ’1)2))
2+(βˆ’1)=1
The integral evaluates to 1, indicating that the total area under the PDF is 1. Therefore, the c value is 1.
b. To find the expected value (E(X)), we need to calculate the mean of the random variable. The expected value is given by the integral of x times the PDF:
E(X)=βˆ«βˆ’βˆžβˆžxΒ·f(x)dx
Using the given PDF:
E(X)=βˆ«βˆ’11xΒ·2|x|dx
Splitting the integral into two parts based on the absolute value:
βˆ«βˆ’11xΒ·2xdx+βˆ«βˆ’11xΒ·βˆ’2xdx
Integrating each part:
[23x3]βˆ’11+[βˆ’23x3]βˆ’11
Evaluating at the limits:
(23(1)3βˆ’23(βˆ’1)3)+(βˆ’23(1)3+23(βˆ’1)3)
43+43=83
Therefore, the expected value (mean) of the random variable X is 83.
c. To find the variance (Var(X)), we need to calculate the second central moment of the random variable. The variance is given by:
Var(X)=E(X2)βˆ’(E(X))2
To calculate E(X2), we need to find the integral of x2 times the PDF:
E(X2)=βˆ«βˆ’βˆžβˆžx2Β·f(x)dx
Using the given PDF:
E(X2)=βˆ«βˆ’11x2Β·2|x|dx
Splitting the integral into two parts based on the absolute value:
βˆ«βˆ’11x2Β·2xdx+βˆ«βˆ’11x2Β·βˆ’2xdx
Integrating each part:
[23x4]βˆ’11+[βˆ’23x4]βˆ’11
Evaluating at the limits:
(23(1)4βˆ’23(βˆ’1)4)+(βˆ’23(1)4+23(βˆ’1)4)
23+23=43
Now we can calculate the variance:
Var(X)=E(X2)βˆ’(E(X))2
Var(X)=43βˆ’(83)2
Var(X)=43βˆ’649
Var(X)=129βˆ’649
Var(X)=βˆ’529
Therefore, the variance of the random variable X is βˆ’529.
d. Finally, to find the probability that X>c, we need to calculate the integral of the PDF from c to ∞:
P(X>c)=∫c∞f(x)dx
Using the given PDF and c=0 in this case:
P(X>0)=∫0∞2|x|dx
Splitting the integral into two parts based on the absolute value:
∫0∞2xdx+∫0βˆžβˆ’2xdx
Integrating each part:
[x2]0∞+[βˆ’x2]0∞
Evaluating at the limits:
(limtβ†’βˆžt2βˆ’02)+(βˆ’limtβ†’βˆžt2βˆ’02)
limtβ†’βˆžt2+limtβ†’βˆž(βˆ’t2)
As t approaches infinity, both limits go to infinity. However, since we have a positive limit and a negative limit, they cancel each other out, resulting in zero.
Therefore, P(X>0)=0.

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