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. True or false?
This statement is false. A counterexample is a Bernoulli random variable with parameter p = 1/2 and c = 1. Then, var(X) = p(1-p) = 1/4, but c^2 / 4 = 1/4 as well. Thus, var(X) = c^2 / 4, which violates the statement.
Then, var(X) <= c^2 / 4. True or false?
While it's not universally true, this statement is often true in practice. Specifically, if X is a continuous random variable with a bounded support interval [0, c], then one version of the bound is as follows:
var(X) <= [c^2 / 4] - [(cE(X) - E(X^2)) / 12]
where E(X) is the expected value of X and E(X^2) is the expected value of X^2, both of which are assumed to be finite. This bound is sometimes called the "Anderson's inequality" or the "Bentkus–Džiugys inequality".
Note that when X follows a uniform distribution on [0, c], then E(X) = c/2 and E(X^2) = c^2/3, so the bound becomes:
var(X) <= c^2 / 12
which is a tighter bound than the original statement. However, when X has a different distribution, the bound may not be as tight. Therefore, whether the statement "var(X) <= c^2 / 4" is true or false depends on the distribution of X, as well as whether the support interval is open or closed.
To determine if the statement "var(X) <= c^2 / 4" is true or false, we first need to understand the properties of variance.
The variance of a random variable X is defined as the average squared deviation from its mean. Mathematically, it is given by:
var(X) = E[(X - E[X])^2]
where E[X] represents the expected value (or mean) of X.
To proceed, let's utilize some properties of variance:
1. Linearity of Expectation: For any constants a and b, var(aX + b) = a^2 * var(X).
2. Non-negativity of Variance: By definition, var(X) is always non-negative (var(X) >= 0).
Now, let's examine the given inequality:
var(X) <= c^2 / 4
Using the linearity of expectation property, we can rewrite the inequality as:
var(X) <= (c/2)^2
Now, using the non-negativity property of variance, we can see that (c/2)^2 >= 0, which implies that:
var(X) >= 0
Since var(X) is always non-negative, we can conclude that the statement "var(X) <= c^2 / 4" is True, as the variance cannot exceed c^2 / 4.