Question 1177403: A degenerate random variable is a random variable taking a constant value. Let X = c. Show that E(X) = c, and Var(X) = 0. Also find the cumulative distribution function of the degenerate distribution of X
Thank you..
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's explore the properties of a degenerate random variable.
**Expected Value (E(X))**
For a discrete random variable X, the expected value is calculated as:
```
E(X) = Σ [x * P(X = x)]
```
where the sum is taken over all possible values of x.
In the case of a degenerate random variable where X = c (a constant), there's only one possible value. Therefore:
```
E(X) = c * P(X = c)
```
Since X always takes the value c, the probability P(X = c) = 1. Hence:
```
E(X) = c * 1 = c
```
**Variance (Var(X))**
The variance of a random variable X is calculated as:
```
Var(X) = E(X^2) - (E(X))^2
```
For our degenerate random variable:
```
E(X^2) = c^2 * P(X = c) = c^2 * 1 = c^2
```
Therefore:
```
Var(X) = E(X^2) - (E(X))^2 = c^2 - c^2 = 0
```
This makes intuitive sense, as a degenerate random variable has no variability.
**Cumulative Distribution Function (CDF)**
The CDF, F(x), of a random variable X is defined as:
```
F(x) = P(X ≤ x)
```
For our degenerate random variable where X = c:
* If x < c, then P(X ≤ x) = 0, since X cannot take any value less than c.
* If x ≥ c, then P(X ≤ x) = 1, since X always takes the value c.
Therefore, the CDF of the degenerate distribution is:
```
F(x) =
0, x < c
1, x ≥ c
```
This CDF is a step function that jumps from 0 to 1 at x = c.
|
|
|
