Question

1. The daily rainfall in Dublin (measured in millimeters) is modelled using a gamma distribution with parameters α = 0.8 and β = 0.4. . (c) Consider the overall rainfall in 365 days, and use mgfs and their properties to prove that this is Ga (292, 0.4). .

Answer 1

Let X_i denotes the rainfall in Dubin (measured in millimeters) of i-th day.

Here,

$X_i \sim Gamma(\alpha = 0.8, \beta = 0.4)$

The probability density function of X is

0.40,8 -0.4x r0.8-1, Γ > 0 f(z) =-0.81-e

The MGF of X is

0.4.40.8-1d 「(0.8) r_.da E(ex)

$\Rightarrow E(e^{tX}) = \frac{0.4^{0.8}}{\Gamma (0.8)}\int_{0}^{\infty }e^{-(0.4-t)x}x^{0.8 - 1}dx$

$\Rightarrow E(e^{tX}) = \frac{0.4^{0.8}}{\Gamma (0.8)}\int_{0}^{\infty }e^{-(0.4-t)x}x^{0.8 - 1}dx$

$\Rightarrow E(e^{tX}) = \frac{0.4^{0.8}}{\Gamma (0.8)}*\frac{\Gamma (0.8)}{(0.4 -t)^{0.8}}$

$\Rightarrow E(e^{tX}) = \frac{0.4^{0.8}}{(0.4 -t)^{0.8}}$

$\Rightarrow E(e^{tX}) = \frac{1}{(1 -\frac{t}{0.4})^{0.8}}$

Let us denote $S = \sum_{i = 1}^{365}X_i$

The MGF of S will be

$M_S(t) = E(e^{tS}) = E(e^{t\sum_{i = 1}^{365}X_i}) =E(e^{tX_1})*E(e^{tX_2})*....*E(e^{tX_{365}})$ since X_i 's are independent.

$\Rightarrow M_S(t) =(\frac{1}{(1 -\frac{t}{0.4})^{0.8}})*(\frac{1}{(1 -\frac{t}{0.4})^{0.8}})*....*(\frac{1}{(1 -\frac{t}{0.4})^{0.8}})$

$\Rightarrow M_S(t) =(\frac{1}{(1 -\frac{t}{0.4})^{0.8}})^{365}$

$\Rightarrow M_S(t) =\frac{1}{(1 -\frac{t}{0.4})^{0.8*365}}$

$\Rightarrow M_S(t) =\frac{1}{(1 -\frac{t}{0.4})^{292}}$

Since MGF uniquely identifies a distribution, so we can say

$S \sim Gamma(\alpha = 292, \beta = 0.4)$