1. The problem gives three different functions defined on different intervals: - $$f(x) = \frac{3x^2}{7^3} e^{-\frac{x^3}{7}}, \quad 0 < x < \infty$$ - $$f(x) = 30x(1 - x)^4, \quad 0 < x < 1$$ - $$f(x) = \frac{1}{\theta} e^{\frac{x - \delta}{\theta}}, \quad \delta < x < \infty$$ 2. These functions appear to be probability density functions (pdfs) because they are defined on intervals and involve exponential or polynomial terms common in pdfs. 3. To verify if each is a valid pdf, we check two conditions: - The function must be non-negative on its domain. - The integral over its domain must equal 1. 4. For the first function: - The function is non-negative since $x^2 \geq 0$ and the exponential is positive. - The integral is $$\int_0^\infty \frac{3x^2}{7^3} e^{-\frac{x^3}{7}} dx$$ Using substitution $t = \frac{x^3}{7}$, then $dt = \frac{3x^2}{7} dx$ so $dx = \frac{7}{3x^2} dt$. Substituting: $$\int_0^\infty \frac{3x^2}{7^3} e^{-t} \frac{7}{3x^2} dt = \int_0^\infty \frac{1}{7^2} e^{-t} dt = \frac{1}{49} \int_0^\infty e^{-t} dt = \frac{1}{49} [ -e^{-t} ]_0^\infty = \frac{1}{49} (1 - 0) = \frac{1}{49}$$ Since this integral is $\frac{1}{49}$, not 1, the function as given is not a valid pdf unless multiplied by 49. 5. For the second function: - The function is non-negative on $0 < x < 1$. - The integral is $$\int_0^1 30x(1-x)^4 dx$$ Using Beta function or binomial expansion: $$\int_0^1 x^1 (1-x)^4 dx = \text{Beta}(2,5) = \frac{\Gamma(2) \Gamma(5)}{\Gamma(7)} = \frac{1! \cdot 4!}{6!} = \frac{24}{720} = \frac{1}{30}$$ Multiplying by 30: $$30 \times \frac{1}{30} = 1$$ So the integral equals 1, confirming it is a valid pdf. 6. For the third function: - The function is non-negative for $x > \delta$. - The integral is $$\int_\delta^\infty \frac{1}{\theta} e^{\frac{x - \delta}{\theta}} dx$$ Rewrite the exponent: $$e^{\frac{x - \delta}{\theta}} = e^{\frac{x}{\theta}} e^{-\frac{\delta}{\theta}}$$ This grows exponentially as $x \to \infty$, so the integral diverges. If the function was meant to be $$f(x) = \frac{1}{\theta} e^{-\frac{x - \delta}{\theta}}, \quad x > \delta$$ then the integral would be $$\int_\delta^\infty \frac{1}{\theta} e^{-\frac{x - \delta}{\theta}} dx = 1$$ which is the pdf of an exponential distribution with location $\delta$ and scale $\theta$. 7. Summary: - First function is not normalized; multiply by 49 to normalize. - Second function is a valid pdf. - Third function as given is not a valid pdf; likely a typo in the exponent sign.