Question

A random variable X has the probability density function (pdf) f(x) given as:

f(x) = { x/c, 1 ≤ x ≤ 2, 0 otherwise }

(a) Find the value of the constant c.
(b) Calculate the cumulative distribution function (cdf) for X.
(c) Determine the variance (Var(X)) of X.
(d) Calculate the variance (Var(lnX)) of the natural logarithm of X.
(e) Compute the probability P(|X - E(X)| ≥ 7/10 * Var(X)).
(f) Utilize the Chebyshev Inequality to provide an upper bound on the probability mentioned in part (e).

Answer 1

The student question pertains to calculating specific properties of a continuous random variable X with a given probability density function, including the constant normalization, cdf, variance, and using the Chebyshev Inequality.

Step-by-step explanation:

The student's question revolves around the properties of a continuous random variable X with a given probability density function (pdf) and finding various statistical measures related to it. We solve this step by step:

1. To find the value for c, we need to use the fact that the integral of the density function over the entire space must equal 1. We solve ∫ f(x) dx = 1 from x=1 to x=2.
2. To calculate the cumulative distribution function (cdf), we integrate the pdf f(x) from 1 to a variable upper limit x.
3. To determine the variance Var(X), we calculate E(X^2) - [E(X)]^2, where E(X) is the expected value of X.
4. For the variance Var(lnX), we need to find the expected value of (lnX) and (lnX)^2 and use these to calculate the variance.
5. The probability P(|X - E(X)| ≥ 7/10 * Var(X)) is calculated directly from the cdf.
6. The Chebyshev Inequality provides an upper bound for the probability that a random variable deviates from its mean by a certain amount.

Answer 2

(a)
`\( c = 1/4 \)`

(b)
`\( F(x) = (1)/(8)x^2 - (1)/(8) \), for \( 1 \leq x \leq 2 \), \( F(x) = 0 \) otherwise.`

(c)
`\( Var(X) = (1)/(40) \)`

(d)
`\( Var(\ln(X)) = (1)/(120) \)`

(e)
`\( P(|X - E(X)| \geq (7)/(10) \cdot Var(X)) = (1)/(6) \)`

(f)
`\( P(|X - E(X)| \geq (7)/(10) \cdot Var(X)) \leq (25)/(36) \)`

Step-by-step explanation:

(a) Constant c : To find c , we integrate f(x) over its entire range, equating it to 1 (since it's a probability density function). The integral of f(x) from 1 to 2 is
`\( \int_(1)^(2) (x)/(c) dx \)`, and setting it equal to 1, we solve for c to get
`\( c = 1/4 \)`.

(b) Cumulative Distribution Function F(x) : Integrating f(x) from 1 to x gives the cumulative distribution function F(x) . For
`\( 1 \leq x \leq 2 \)`, F(x) =
`(1)/(8)x^2 - (1)/(8) \), and \( F(x) = 0 \)` otherwise.

(c) Variance
`\( Var(X) \)`: The variance of a random variable X is given by
`\( Var(X) = E(X^2) - [E(X)]^2 \)`. We find E(X) and
`\( E(X^2) \)` by integrating
`\( x \cdot f(x) \)` and
`\( x^2 \cdot f(x) \)`, respectively. The variance
`\( Var(X) \)` is then calculated.

(d) Variance of
`\( \ln(X) \)`: We use the property that
`\( Var(\ln(X)) = [Var(X)] / [E(X)]^2 \)`. Calculating
`\( Var(\ln(X)) \)` involves finding
`\( E(\ln(X)) \)` and
`\( E[(\ln(X))^2] \)`, using the properties of logarithmic transformations.

(e) Probability
`\( P(|X - E(X)| \geq (7)/(10) \cdot Var(X)) \)`: This involves computing the probability that X deviates from its expected value by a certain proportion of its variance.

(f) Chebyshev Inequality: Applying Chebyshev's Inequality provides an upper bound for the probability in part (e). It is given by
`\( P(|X - E(X)| \geq (7)/(10) \cdot Var(X)) \leq (1)/(k^2) \)`, where k is the proportionality factor. For this problem,
`\( k = (10)/(7) \)`, and the upper bound is calculated as
`\( (25)/(36) \)`.