Here is a simple way to create a random variable X that has mean μ and stan-
dard deviation σ: X takes only the two values μ−σ and μ+σ, eachwith probability
0.5. Use the definition of the mean and variance for discrete random
variables to show that X does have mean μ and standard deviation σ.
To show that the random variable X has a mean of μ and a standard deviation of σ, we will use the definition of the mean and variance for discrete random variables.
1. Mean (Expected Value):
The mean (μ) of a discrete random variable X is calculated by multiplying each possible value of X by its corresponding probability and summing them up.
For our random variable X, we have two possible values: μ-σ and μ+σ, each with a probability of 0.5:
E(X) = (μ-σ)(0.5) + (μ+σ)(0.5)
= (0.5)(μ-σ + μ+σ)
= (0.5)(2μ)
= μ
Therefore, the mean of X is indeed μ.
2. Variance:
The variance (σ^2) of a discrete random variable X is calculated by subtracting the mean from each possible value of X, squaring the result, multiplying it by its corresponding probability, and summing them up.
For our random variable X, the squared differences from the mean are (μ-σ - μ)^2 and (μ+σ - μ)^2:
Var(X) = (μ-σ - μ)^2(0.5) + (μ+σ - μ)^2(0.5)
= (-σ)^2(0.5) + σ^2(0.5)
= σ^2(0.5) + σ^2(0.5)
= 0.5σ^2 + 0.5σ^2
= σ^2
Therefore, the variance of X is indeed σ^2.
3. Standard Deviation:
The standard deviation (σ) is the square root of the variance.
Standard Deviation (σ) = √[Var(X)]
= √[σ^2]
= σ
Therefore, the standard deviation of X is indeed σ.
In conclusion, we have shown that the random variable X, which takes the values μ-σ and μ+σ with equal probabilities of 0.5, has a mean of μ and a standard deviation of σ, using the definitions of the mean and variance for discrete random variables.
To show that the random variable X has a mean of μ and a standard deviation of σ, we need to calculate the mean and variance of X using the definition for discrete random variables.
Let's start by calculating the mean (also known as the expected value) of X. The formula for the mean of a discrete random variable is given by:
E(X) = Σ(xi * P(xi))
Where xi represents the possible values of X and P(xi) represents the probability of X taking the value xi.
In this case, X only takes the two values μ-σ and μ+σ, each with a probability of 0.5. Therefore, we can plug in these values into the formula:
E(X) = (μ-σ * 0.5) + (μ+σ * 0.5)
= (0.5μ - 0.5σ) + (0.5μ + 0.5σ)
= 0.5μ - 0.5σ + 0.5μ + 0.5σ
= μ
As you can see, the mean E(X) equals μ, which confirms that X has a mean of μ.
Next, let's calculate the variance of X. The formula for the variance of a discrete random variable is given by:
Var(X) = Σ(xi - E(X))^2 * P(xi)
Using the values of X and their probabilities, we can plug them into the formula:
Var(X) = [(μ-σ - μ)^2 * 0.5] + [(μ+σ - μ)^2 * 0.5]
= [(-σ)^2 * 0.5] + [(σ)^2 * 0.5]
= [σ^2 * 0.5] + [σ^2 * 0.5]
= σ^2 * (0.5 + 0.5)
= σ^2
As you can see, the variance Var(X) equals σ^2, which confirms that X has a standard deviation of σ.
Therefore, based on the calculations above, we have shown that the random variable X created using the given method does have a mean of μ and a standard deviation of σ.