Question

The probability density function of a continuous random variable X is

• A. $\frac{7}{4}$
• B. $\frac{43}{20}$
• C. $\frac{15}{8}$
• D. $\frac{12}{7}$

Hint:

• Mean of a continuous random variable: $E[X] = \int_{-\infty}^{\infty} x f(x) dx$.
• If $f(x)$ is piecewise, split the integral into corresponding parts.
• For $\int x(a-x)^n dx$, substitution $u=a-x$ is often helpful.
• Basic integration: $\int x^k dx = x^{k+1}/(k+1)$.
• Add fractions carefully using a common denominator.
• Before starting, it's good practice to verify $\int_{-\infty}^{\infty} f(x) dx = 1$. $\int_0^1 x^3 dx = [x^4/4]_0^1 = 1/4$. $\int_1^2 (2-x)^3 dx$. Let $u=2-x, du=-dx$. $\int_1^0 u^3(-du) = \int_0^1 u^3 du = [u^4/4]_0^1 = 1/4$. $\int_2^3 (1/2) dx = [x/2]_2^3 = 3/2 - 2/2 = 1/2$. Total probability = $1/4 + 1/4 + 1/2 = 1/2 + 1/2 = 1$. The pdf is valid.

Answer 1

The Correct Option is A

Given the probability density function (pdf) of a continuous random variable \(X\):

\[ f(x) = \begin{cases} x^3, & 0 \leq x \leq 1 \\ (2 - x)^3, & 1 \leq x \leq 2 \\ \frac{1}{2}, & 2 \leq x \leq 3 \\ 0, & \text{otherwise} \end{cases} \]

Step 1: Verify that \( f(x) \) is a valid pdf

We should confirm \(\int_{-\infty}^\infty f(x) dx = 1\). Given piecewise nature:

\[ \int_0^1 x^3 dx + \int_1^2 (2 - x)^3 dx + \int_2^3 \frac{1}{2} dx = 1 \]

Check if true:

• \(\int_0^1 x^3 dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4}\)
• \(\int_1^2 (2 - x)^3 dx\), substitute \( t = 2 - x \Rightarrow dt = -dx \), when \( x=1, t=1 \), when \( x=2, t=0 \): \[ \int_1^2 (2 - x)^3 dx = \int_1^0 t^3 (-dt) = \int_0^1 t^3 dt = \frac{1}{4} \]
• \(\int_2^3 \frac{1}{2} dx = \frac{1}{2} (3 - 2) = \frac{1}{2}\)

Total:

\[ \frac{1}{4} + \frac{1}{4} + \frac{1}{2} = 1 \]

So \( f(x) \) is a valid pdf.

Step 2: Find the mean \( \mu = E[X] = \int x f(x) dx \)

Calculate piecewise:

\[ E[X] = \int_0^1 x \cdot x^3 dx + \int_1^2 x \cdot (2 - x)^3 dx + \int_2^3 x \cdot \frac{1}{2} dx \]

Calculate each integral:

1. \[ \int_0^1 x^4 dx = \left[ \frac{x^5}{5} \right]_0^1 = \frac{1}{5} \]
2. For \(\int_1^2 x (2 - x)^3 dx\), substitute \( t = 2 - x \Rightarrow x = 2 - t \), and \( dt = -dx \): \[ \int_1^2 x (2 - x)^3 dx = \int_1^0 (2 - t) t^3 (-dt) = \int_0^1 (2 - t) t^3 dt = \int_0^1 (2 t^3 - t^4) dt \] \[ = 2 \int_0^1 t^3 dt - \int_0^1 t^4 dt = 2 \cdot \frac{1}{4} - \frac{1}{5} = \frac{1}{2} - \frac{1}{5} = \frac{5}{10} - \frac{2}{10} = \frac{3}{10} \]
3. \[ \int_2^3 \frac{x}{2} dx = \frac{1}{2} \left[ \frac{x^2}{2} \right]_2^3 = \frac{1}{2} \left( \frac{9}{2} - \frac{4}{2} \right) = \frac{1}{2} \times \frac{5}{2} = \frac{5}{4} \]

Step 3: Add the results

\[ E[X] = \frac{1}{5} + \frac{3}{10} + \frac{5}{4} = \frac{2}{10} + \frac{3}{10} + \frac{12.5}{10} = \frac{17.5}{10} = \frac{35}{20} = \frac{7}{4} \]

Answer: The mean of the random variable is \(\boxed{\frac{7}{4}}\).