Question 1203805: Sometimes it is desirable to assign numerical "code" values to experimental responses that are not basically of numerical type For example in testing the color preferences of experimental subjects suppose that the colors blue, green, and red occur with probabilities 1/4, 1/4, and 1/2, respectively. A different integer value is assigned to each color, and this corresponds to a random variable X that can take on one of these three integer values.
(a) Can f(x) = (1/4)^|x|(1/2)^1-|x| for x = -1, 1, 0 be used as a pdf for this experiment?
(b) Can f(x) = (2 x)(1/2)^2 for x = 0, 1, 2 be used?
(c) Can f(x)=(1-x)/4 for x= -1,0,2 be used?
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! **To determine if a function can be used as a probability density function (pdf) for a discrete random variable, it must satisfy the following conditions:**
1. **Non-negativity:** f(x) ≥ 0 for all possible values of x.
2. **Normalization:** The sum of probabilities for all possible values of x must equal 1.
Let's analyze each of the given functions:
**a) f(x) = (1/4)^|x| * (1/2)^(1-|x|) for x = -1, 0, 1**
* **Calculate f(x) for each value of x:**
* f(-1) = (1/4)^|-1| * (1/2)^(1-|-1|) = (1/4) * (1/2)^0 = 1/4
* f(0) = (1/4)^|0| * (1/2)^(1-|0|) = 1 * (1/2)^1 = 1/2
* f(1) = (1/4)^|1| * (1/2)^(1-|1|) = (1/4) * (1/2)^0 = 1/4
* **Check conditions:**
* Non-negativity: All values of f(x) are positive.
* Normalization: f(-1) + f(0) + f(1) = 1/4 + 1/2 + 1/4 = 1
* **Conclusion:** This function **can** be used as a pdf for the experiment.
**b) f(x) = (2x)(1/2)^2 for x = 0, 1, 2**
* **Calculate f(x) for each value of x:**
* f(0) = (2 * 0)(1/2)^2 = 0
* f(1) = (2 * 1)(1/2)^2 = 1/2
* f(2) = (2 * 2)(1/2)^2 = 1
* **Check conditions:**
* Non-negativity: f(0) is 0, which does not violate the non-negativity condition.
* Normalization: f(0) + f(1) + f(2) = 0 + 1/2 + 1 = 3/2 ≠ 1
* **Conclusion:** This function **cannot** be used as a pdf because it does not satisfy the normalization condition.
**c) f(x) = (1-x)/4 for x = -1, 0, 2**
* **Calculate f(x) for each value of x:**
* f(-1) = (1 - (-1))/4 = 1/2
* f(0) = (1 - 0)/4 = 1/4
* f(2) = (1 - 2)/4 = -1/4
* **Check conditions:**
* Non-negativity: f(2) is negative, violating the non-negativity condition.
* **Conclusion:** This function **cannot** be used as a pdf because it does not satisfy the non-negativity condition.
**Summary:**
* **Only the function in part (a) can be used as a pdf for the given experiment.**
* The other two functions either violate the non-negativity condition or the normalization condition, which are essential requirements for a valid probability density function.
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