To determine whether each statement is true or false, we can break down the statements and analyze their components.
1. Statement: Let X be a random variable that takes values between 0 and c only, for some c≥0, so that P(0≤X≤c)=1. Then, var(X)≤c^2/4.
To determine the truth of this statement, we need to understand the concepts of random variables and variance.
- A random variable is a variable that can take on different values based on the outcomes of a random process or experiment.
- Variance is a measure of how spread out the values of a random variable are around the mean.
In this statement, X is a random variable that can take values between 0 and c only, with a probability of 1. This means that X is confined to this range and will always fall within it.
The variance of X (denoted as var(X)) is defined as the average of the squared differences of each value of X from its mean.
To determine if var(X)≤c^2/4, we first need to find the mean of X. Since X can only take values between 0 and c, the mean (denoted as μ) will be the average of these values.
Therefore, μ = (0 + c)/2 = c/2.
Next, we compute the variance using the formula:
var(X) = E[(X - μ)^2],
where E represents the expectation or average.
For X to take values between 0 and c with a probability of 1, the probability distribution function fX(x) will be a constant between 0 and c. Therefore, fX(x) = 1/c for 0≤ x ≤ c, and 0 otherwise.
Plugging these values into the expectation formula, we get:
var(X) = E[(X - c/2)^2]
= ∫[0,c] (x - c/2)^2 * (1/c) dx
= 1/c ∫[0,c] (x^2 - cx + c^2/4) dx
= 1/c [(x^3/3 - c/2 * x^2 + c^2/4 * x)] ∣[0,c]
= 1/c [(c^3/3 - c/2 * c^2 + c^2/4 * c) - (0)]
= 1/c [(c^3/3 - c^3/2 + c^3/4)]
= 1/c * (c^3/12)
= c^2/12.
Now we compare c^2/12 with c^2/4:
c^2/12 ≤ c^2/4,
Simplifying, we see that:
1/12 ≤ 1/4,
Since 1/12 is indeed less than or equal to 1/4, we can conclude that var(X) ≤ c^2/4 is true. Therefore, the statement is true.
2. Statement: X and Y are continuous random variables. If X∼N(μ,σ^2), Y=aX+b, and a>0, then Y∼N(aμ+b,a^2σ^2).
To determine the truth of this statement, we need to understand the concept of transformations of random variables and the properties of normal distributions.
- In this statement, X is a normally distributed random variable with mean μ and variance σ^2 (denoted as X∼N(μ,σ^2)).
- Y is a new random variable formed by transforming X, where Y=aX+b. Here, a is a scaling factor and b is a shift factor.
To find the distribution of Y, we need to determine its mean and variance.
The mean of Y (denoted as μY) can be found by substituting the transformation into the mean of X:
μY = aμ + b.
The variance of Y (denoted as var(Y)) can be found using the properties of variances and the transformation of variables:
var(Y) = a^2 * var(X).
From the given statement, it claims that Y∼N(aμ+b,a^2σ^2).
To verify the truth of this claim, we compare the derived mean and variance of Y with the claimed mean and variance.
In the statement, it is stated that a>0 which implies that a is positive. However, the derived variance of Y is a^2 * var(X). Therefore, the given statement that Y∼N(aμ+b,a^2σ^2) is false, as it should be Y∼N(aμ+b,|a|σ^2) instead.
3. Statement: The expected value of a non-negative continuous random variable X, which is defined by E[X]=∫∞0 xfX(x)dx, also satisfies E[X]=∫∞0 P(X>t)dt.
To determine the truth of this statement, we need to understand the concept of expected value and the connection between expected value and the survival function of a random variable.
- The expected value (denoted as E[X]) is a measure of the center or average of a random variable X. It represents the weighted sum of all possible values of X, with the weights given by their probabilities.
In this statement, the expected value of X is defined using the integral form:
E[X] = ∫∞0 xfX(x)dx,
where fX(x) is the probability density function of X.
The survival function of a random variable X is defined as the probability that X is greater than some value t:
P(X > t).
To compare the expected value with the survival function, we can relate it to the concept of integration by parts.
Using integration by parts, we can rewrite the integral in the statement as:
∫∞0 xfX(x)dx = -∫∞0 x d[P(X > x)].
Applying integration by parts, we get:
∫∞0 xfX(x)dx = -[x * P(X > x) ∣[0,∞] - ∫∞0 P(X > x)dx].
The term [x * P(X > x) ∣[0,∞]] evaluates to 0 when evaluated at both 0 and ∞, as P(X > ∞) = 0.
Therefore, the integral becomes:
∫∞0 xfX(x)dx = -∫∞0 P(X > x)dx.
By convention, we switch the sign to make it positive:
∫∞0 xfX(x)dx = ∫∞0 P(X > x)dx.
Comparing this result with the statement, we can see that E[X] = ∫∞0 P(X > t)dt is indeed true. Therefore, the statement is true.