1. **State the problem:** Find the constant $c$ for the given PDF $f(x) = c x e^{-2x}, x > 0$, with total probability 1. 2. **Write the total probability condition:** $$\int_0^{\infty} c x e^{-2x} dx = 1$$ 3. **Calculate the integral:** Using integration by parts or the gamma function formula, $$\int_0^{\infty} x e^{-2x} dx = \frac{1}{2^2} = \frac{1}{4}$$ 4. **Solve for $c$:** Substitute: $$c \times \frac{1}{4} = 1 \implies c = 4$$ 5. **Find the harmonic mean (HM) of $X$:** The harmonic mean is: $$HM = \frac{1}{E\left(\frac{1}{X}\right)}$$ Calculate $E\left(\frac{1}{X}\right)$: $$E\left(\frac{1}{X}\right) = \int_0^{\infty} \frac{1}{x} \cdot 4 x e^{-2x} dx = 4 \int_0^{\infty} e^{-2x} dx = 4 \times \frac{1}{2} = 2$$ Therefore: $$HM = \frac{1}{2}$$ 6. **Find the moment generating function (m.g.f.) of $X$:** By definition: $$M_X(t) = E(e^{tX}) = \int_0^{\infty} e^{t x} 4 x e^{-2x} dx = 4 \int_0^{\infty} x e^{-(2 - t)x} dx$$ This integral converges if $t < 2$. Recall: $$\int_0^{\infty} x e^{-a x} dx = \frac{1}{a^2}$$ Thus: $$M_X(t) = \frac{4}{(2 - t)^2}, \quad t < 2$$ 7. **Find the m.g.f. of $Y = \frac{1}{X} + 9$:** Moment generating function: $$M_Y(t) = E(e^{tY}) = E\left(e^{t\left(\frac{1}{X} + 9\right)}\right) = e^{9t} E\left(e^{\frac{t}{X}}\right)$$ Calculate: $$E\left(e^{\frac{t}{X}}\right) = \int_0^{\infty} e^{\frac{t}{x}} 4 x e^{-2x} dx$$ This integral does not simplify to elementary functions. Hence, $$M_Y(t) = e^{9t} \int_0^{\infty} 4 x e^{\frac{t}{x}} e^{-2x} dx$$ 8. **Find the second factorial moment $E[X(X-1)]$:** Recall: $$E[X(X-1)] = E[X^2] - E[X]$$ Calculate $E[X]$: $$E[X] = \int_0^{\infty} x 4 x e^{-2x} dx = 4 \int_0^{\infty} x^2 e^{-2x} dx$$ Using gamma function: $$\int_0^{\infty} x^n e^{-a x} dx = \frac{n!}{a^{n+1}}$$ So: $$E[X] = 4 \times \frac{2!}{2^3} = 4 \times \frac{2}{8} = 1$$ Calculate $E[X^2]$: $$E[X^2] = 4 \int_0^{\infty} x^3 e^{-2x} dx = 4 \times \frac{3!}{2^4} = 4 \times \frac{6}{16} = \frac{24}{16} = 1.5$$ Thus: $$E[X(X-1)] = 1.5 - 1 = 0.5$$ 9. **Find $P(X < 3)$:** Calculate the CDF: $$F_X(3) = \int_0^3 4 x e^{-2x} dx$$ Integrate by parts: $$\int x e^{-2x} dx = -\frac{x}{2} e^{-2x} - \frac{1}{4} e^{-2x} + C$$ So: $$\int_0^3 4 x e^{-2x} dx = 4 \left[-\frac{x}{2} e^{-2x} - \frac{1}{4} e^{-2x}\right]_0^3 = 4 \left[-\frac{3}{2} e^{-6} - \frac{1}{4} e^{-6} + \frac{1}{4} \right]$$ Simplify: $$= 4 \left[-\frac{7}{4} e^{-6} + \frac{1}{4} \right] = 4 \times \frac{1 - 7 e^{-6}}{4} = 1 - 7 e^{-6}$$ 10. **Find the second non-central moment around the point 3, $E[(X-3)^2]$:** Use identity: $$E[(X-3)^2] = E[X^2] - 2 \times 3 \times E[X] + 3^2$$ Using previous values: $$E[X^2] = 1.5, \quad E[X] = 1$$ Thus: $$E[(X-3)^2] = 1.5 - 6 + 9 = 4.5$$