Probability Density Function Validity Analysis

A function $f(x)$ on the real line is considered a valid probability density function (PDF) if it satisfies two key conditions:

1. Non-negativity: $f(x) \ge 0$ for all real values of $x$.
2. Normalization: The integral of $f(x)$ over the entire real line must be equal to 1. That is, $\int_{-\infty}^{\infty} f(x) dx = 1$.

We are given that $f(x)$ is a valid probability density on the real line. This means $f(x) \ge 0$ for all $x$ and $\int_{-\infty}^{\infty} f(x) dx = 1$. We need to check which of the given options is NOT a valid probability density by examining if they satisfy these two conditions.

Analyzing Option 1: $g(x) = f(x + 1)$

• Non-negativity: Since $f(y) \ge 0$ for any real number $y$, $f(x+1)$ will also be $\ge 0$ for all real $x$. This condition is satisfied.
• Normalization: We need to evaluate $\int_{-\infty}^{\infty} f(x+1) dx$. Let $y = x+1$. Then $dy = dx$. The limits of integration remain the same: as $x \to -\infty$, $y \to -\infty$; as $x \to \infty$, $y \to \infty$.
  The integral becomes $\int_{-\infty}^{\infty} f(y) dy$. Since $f(y)$ is a valid PDF, this integral equals 1.
  Thus, $\int_{-\infty}^{\infty} f(x+1) dx = 1$. This condition is satisfied.

Option 1 is a valid probability density.

Analyzing Option 2: $g(x) = f(2x)$

• Non-negativity: Since $f(y) \ge 0$ for any real number $y$, $f(2x)$ will also be $\ge 0$ for all real $x$. This condition is satisfied.
• Normalization: We need to evaluate $\int_{-\infty}^{\infty} f(2x) dx$. Let $y = 2x$. Then $dy = 2 dx$, which means $dx = \frac{1}{2} dy$. The limits of integration remain the same: as $x \to -\infty$, $y \to -\infty$; as $x \to \infty$, $y \to \infty$.
  The integral becomes $\int_{-\infty}^{\infty} f(y) \frac{1}{2} dy = \frac{1}{2} \int_{-\infty}^{\infty} f(y) dy$. Since $f(y)$ is a valid PDF, $\int_{-\infty}^{\infty} f(y) dy = 1$.
  Thus, $\int_{-\infty}^{\infty} f(2x) dx = \frac{1}{2} \times 1 = \frac{1}{2}$. This integral is not equal to 1. This condition is NOT satisfied.

Option 2 is NOT a valid probability density because it does not integrate to 1.

Analyzing Option 3: $g(x) = 2f(2x - 1)$

• Non-negativity: Since $f(y) \ge 0$, $2f(2x-1)$ will also be $\ge 0$. This condition is satisfied.
• Normalization: We need to evaluate $\int_{-\infty}^{\infty} 2f(2x - 1) dx$. Let $y = 2x - 1$. Then $dy = 2 dx$, which means $dx = \frac{1}{2} dy$. The limits of integration remain the same.
  The integral becomes $\int_{-\infty}^{\infty} 2 f(y) \frac{1}{2} dy = \int_{-\infty}^{\infty} f(y) dy$. Since $f(y)$ is a valid PDF, this integral equals 1.
  Thus, $\int_{-\infty}^{\infty} 2f(2x - 1) dx = 1$. This condition is satisfied.

Option 3 is a valid probability density.

Analyzing Option 4: $g(x) = 3x^2f(x^3)$

• Non-negativity: Since $f(y) \ge 0$, $f(x^3) \ge 0$. Also, $3x^2 \ge 0$ for all real $x$. The product $3x^2f(x^3)$ is therefore $\ge 0$. This condition is satisfied.
• Normalization: We need to evaluate $\int_{-\infty}^{\infty} 3x^2f(x^3) dx$. Let $y = x^3$. Then $dy = 3x^2 dx$. The limits of integration remain the same: as $x \to -\infty$, $y = (-\infty)^3 \to -\infty$; as $x \to \infty$, $y = (\infty)^3 \to \infty$.
  The integral becomes $\int_{-\infty}^{\infty} f(y) dy$. Since $f(y)$ is a valid PDF, this integral equals 1.
  Thus, $\int_{-\infty}^{\infty} 3x^2f(x^3) dx = 1$. This condition is satisfied.

Option 4 is a valid probability density.

Out of the given options, only $f(2x)$ fails the normalization condition, integrating to $\frac{1}{2}$ instead of 1. Therefore, $f(2x)$ is NOT a valid probability density function.