1. **Problem statement:** We have a discrete random variable $Y$ with probability generating function (PGF) $$G_Y(t) = 0.09t^2 + 0.24t^3 + 0.34t^4 + 0.24t^5 + 0.09t^6.$$ We need to find (i) the mean and variance of $Y$, (ii) the PGF of $X$ where $Y$ is the sum of two independent observations of $X$, and (iii) the value of $P(X=2)$. 2. **Recall:** - The mean of a discrete random variable with PGF $G(t)$ is $E[Y] = G'(1)$. - The variance is $Var(Y) = G''(1) + G'(1) - (G'(1))^2$. - If $Y = X_1 + X_2$ where $X_1, X_2$ are independent and identically distributed as $X$, then $G_Y(t) = (G_X(t))^2$. --- ### (i) Find mean and variance of $Y$: 3. Compute $G_Y'(t)$: $$G_Y'(t) = 0.09 \cdot 2 t^{1} + 0.24 \cdot 3 t^{2} + 0.34 \cdot 4 t^{3} + 0.24 \cdot 5 t^{4} + 0.09 \cdot 6 t^{5} = 0.18 t + 0.72 t^{2} + 1.36 t^{3} + 1.2 t^{4} + 0.54 t^{5}.$$ Evaluate at $t=1$: $$G_Y'(1) = 0.18 + 0.72 + 1.36 + 1.2 + 0.54 = 4.0.$$ So, $E[Y] = 4.0$. 4. Compute $G_Y''(t)$: $$G_Y''(t) = 0.18 \cdot 1 + 0.72 \cdot 2 t + 1.36 \cdot 3 t^{2} + 1.2 \cdot 4 t^{3} + 0.54 \cdot 5 t^{4} = 0.18 + 1.44 t + 4.08 t^{2} + 4.8 t^{3} + 2.7 t^{4}.$$ Evaluate at $t=1$: $$G_Y''(1) = 0.18 + 1.44 + 4.08 + 4.8 + 2.7 = 13.2.$$ 5. Calculate variance: $$Var(Y) = G_Y''(1) + G_Y'(1) - (G_Y'(1))^2 = 13.2 + 4.0 - 4.0^2 = 13.2 + 4.0 - 16 = 1.2.$$ --- ### (ii) Find the PGF of $X$: 6. Since $Y = X_1 + X_2$ with independent $X_i$, $$G_Y(t) = (G_X(t))^2.$$ Therefore, $$G_X(t) = \sqrt{G_Y(t)}.$$ 7. Given $G_Y(t)$ is a polynomial of degree 6, and $G_X(t)$ must be a cubic polynomial: Assume $$G_X(t) = a t^2 + b t^3 + c t^4$$ would be too high degree, so instead try $$G_X(t) = p t^1 + q t^2 + r t^3$$ since $X$ takes values 1, 2, 3 (to match degrees). 8. Square $G_X(t)$: $$(p t + q t^2 + r t^3)^2 = p^2 t^2 + 2 p q t^3 + (q^2 + 2 p r) t^4 + 2 q r t^5 + r^2 t^6.$$ 9. Match coefficients with $G_Y(t)$: - $t^2$: $p^2 = 0.09$ so $p = 0.3$ (positive since probabilities positive) - $t^3$: $2 p q = 0.24$ so $q = 0.24 / (2 imes 0.3) = 0.4$ - $t^4$: $q^2 + 2 p r = 0.34$ so $0.4^2 + 2 imes 0.3 imes r = 0.34$ $$0.16 + 0.6 r = 0.34 ightarrow 0.6 r = 0.18 ightarrow r = 0.3$$ - $t^5$: $2 q r = 0.24$ check: $2 imes 0.4 imes 0.3 = 0.24$ correct - $t^6$: $r^2 = 0.09$ so $r = 0.3$ consistent 10. So, $$G_X(t) = 0.3 t + 0.4 t^2 + 0.3 t^3.$$ --- ### (iii) Find $P(X=2)$: 11. The coefficient of $t^2$ in $G_X(t)$ is $P(X=2)$, so $$P(X=2) = 0.4.$$ --- **Final answers:** - Mean of $Y$ is $4.0$. - Variance of $Y$ is $1.2$. - PGF of $X$ is $G_X(t) = 0.3 t + 0.4 t^2 + 0.3 t^3$. - $P(X=2) = 0.4$.