Question

The variance for continuous probability function \(f(x) = x^2 e^{-x}\) when \(x \ge 0\) is

• A. 1
• B. 0
• C. 3
• D. 2

Hint:

Gamma Distribution. Pdf: \( f(x; \alpha, \beta) = \frac{\beta^\alpha{\Gamma(\alpha) x^{\alpha-1 e^{-\beta x \) for \(x \ge 0\). Mean \(E[X] = \alpha/\beta\). Variance \(Var(X) = \alpha/\beta^2\). Remember \(\Gamma(z) = (z-1)!\) for integer z. First normalize the given function if needed.

Answer 1

The Correct Option is C

First, we need to check if \(f(x)\) is a valid probability density function (pdf)

It must satisfy \(f(x) \ge 0\) for all x (which is true for \(x \ge 0\)) and \( \int_{-\infty}^{\infty} f(x) dx = 1 \)

The integral is \( \int_{0}^{\infty} x^2 e^{-x} dx \)

This is the Gamma function \(\Gamma(z) = \int_0^\infty t^{z-1}e^{-t} dt\)

Here, \(z-1=2\), so \(z=3\)

\( \int_{0}^{\infty} x^2 e^{-x} dx = \Gamma(3) = (3-1)! = 2! = 2 \)

Since the integral is 2, not 1, the given \(f(x)\) is not a standard pdf

However, it is proportional to the Gamma distribution pdf with shape \(\alpha=3\) and rate \(\beta=1\)

The pdf for Gamma(\(\alpha, \beta\)) is \( \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x} \)

Here, \(\alpha=3, \beta=1\), so the pdf is \( p(x) = \frac{1^3}{\Gamma(3)} x^{3-1} e^{-1x} = \frac{1}{2} x^2 e^{-x} \)

So, \( f(x) = 2p(x) \)

Assuming the question intended \(p(x) = \frac{1}{2} x^2 e^{-x}\) to be the pdf: The mean (\(\mu\)) of Gamma(\(\alpha, \beta\)) is \(\alpha/\beta\)

The variance (\(\sigma^2\)) of Gamma(\(\alpha, \beta\)) is \(\alpha/\beta^2\)

For \(p(x)\) with \(\alpha=3, \beta=1\): Mean \(\mu = 3/1 = 3\)

Variance \(\sigma^2 = 3/1^2 = 3\)

Alternatively, calculate moments directly using \(f(x)\) and normalize later, or calculate using \(p(x)\)

\(E[X] = \int_0^\infty x p(x) dx = \int_0^\infty x (\frac{1}{2} x^2 e^{-x}) dx = \frac{1}{2} \int_0^\infty x^3 e^{-x} dx = \frac{1}{2} \Gamma(4) =\frac{1}{2} (3!) = \frac{6}{2} = 3\)\(E[X^2]= \int_0^\infty x^2 p(x) dx = \int_0^\infty x^2 (\frac{1}{2} x^2 e^{-x}) dx = \frac{1}{2} \int_0^\infty x^4 e^{-x} dx = \frac{1}{2} \Gamma(5) = \frac{1}{2} (4!) = \frac{24}{2} = 12\)

Variance \(Var(X) = E[X^2] - (E[X])^2 = 12 - (3)^2 = 12 - 9 = 3\)

The variance is (C).