13.3 Problems on transform methods

Applied probability Page 1 / 2

Calculate directly the generating function $g_{X} (s)$ for the geometric $(p)$ distribution.

g_{X} (s) = E [s^{X}] = \sum_{k = 0}^{\infty} p_{k} s^{k} = p \sum_{k = 0}^{\infty} q^{k} s^{k} = \frac{p}{1 - q s} (geometric series)

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Calculate directly the generating function $g_{X} (s)$ for the Poisson $(μ)$ distribution.

g_{X} (s) = E [s^{X}] = \sum_{k = 0}^{\infty} p_{k} s^{k} = e^{- μ} \sum_{k = 0}^{\infty} \frac{μ^{k} s^{k}}{k!} = e^{- μ} e^{μ s} = e^{μ (s - 1)}

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A projection bulb has life (in hours) represented by $X \sim$ exponential (1/50). The unit will be replaced immediately upon failure or at 60 hours, whichever comes first.Determine the moment generating function for the time Y to replacement.

Y = I_{[0, a]} (X) X + I_{(a, \infty)} (X) a e^{s Y} = I_{[0, a]} (X) e^{s X} + I_{(a, \infty)} (X) e^{a s}

M_{Y} (s) = \int_{0}^{a} e^{s t} λ e^{- λ t} d t + e^{s a} \int_{a}^{\infty} λ e^{- λ t} d t

= \frac{λ}{λ - s} [1 - e^{- (λ - s) a}] + e^{- (λ - s) a}

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Simple random variable X has distribution

X = [- 3 - 2 0 1 4] P X = [0.15 0.20 0.30 0.25 0.10]

Determine the moment generating function for X .
Show by direct calculation the $M_{X}^{'} (0) = E [X]$ and $M_{X}^{''} (0) = E [X^{2}]$ .

M_{X} (s) = 0.15 e^{- 3 s} + 0.20 e^{- 2 s} + 0.30 + 0.25 e^{s} + 0.10 e^{4 s}

M_{X}^{'} (s) = - 3 \cdot 0.15 e^{- 3 s} - 2 \cdot 0.20 e^{- 2 s} + 0 + 0.25 e^{s} + 4 \cdot 0.10 e^{4 s}

M_{X}^{''} (s) = {(- 3)}^{2} \cdot 0.15 e^{- 3 s} + {(- 2)}^{2} \cdot 0.20 e^{- 2 s} + 0 + 0.25 e^{s} + 4^{2} \cdot 0.10 e^{4 s}

Setting $s = 0$ and using $e^{0} = 1$ give the desired results.

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Use the moment generating function to obtain the variances for the following distributions

Exponential $(λ)$ Gamma $(α, λ)$ Normal $(μ, σ^{2})$

Exponential:
$M_{X} (s) = \frac{λ}{λ - s} M_{X}^{'} (s) = \frac{λ}{{(λ - s)}^{2}} M_{X}^{''} (s) = \frac{2 λ}{{(λ - s)}^{3}}$

$E [X] = \frac{λ}{λ^{2}} = \frac{1}{λ} E [X^{2}] = \frac{2 λ}{λ^{3}} = \frac{2}{λ^{2}} Var [X] = \frac{2}{λ^{2}} - {(\frac{1}{λ})}^{2} = \frac{1}{λ^{2}}$
Gamma $(α, λ)$ :
$M_{X} (s) = {(\frac{λ}{λ - s})}^{α} M_{X}^{'} (s) = α {(\frac{λ}{λ - s})}^{α - 1} \frac{λ}{{(λ - s)}^{2}} = α {(\frac{λ}{λ - s})}^{α} \frac{1}{λ - s}$

$M_{X}^{''} (s) = α^{2} {(\frac{λ}{λ - s})}^{α} \frac{1}{λ - s} \frac{1}{λ - s} + α {(\frac{λ}{λ - s})}^{α} \frac{1}{{(λ - s)}^{2}}$

$E [X] = \frac{α}{λ} E [X^{2}] = \frac{α^{2} + α}{λ^{2}} Var [X] = \frac{α}{λ^{2}}$
Normal $(μ, σ)$ :
$M_{X} (s) = exp (\frac{σ^{2} s^{2}}{2} + μ s) M_{X}^{'} (s) = M_{X} (s) \cdot (σ^{2} s + μ)$

$M_{X}^{''} (s) = M_{X} (s) \cdot {(σ^{2} s + μ)}^{2} + M_{X} (s) σ^{2}$

$E [X] = μ E [X^{2}] = μ^{2} + σ^{2} Var [X] = σ^{2}$

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The pair ${X, Y}$ is iid with common moment generating function $\frac{λ^{3}}{{(λ - s)}^{3}}$ . Determine the moment generating function for $Z = 2 X - 4 Y + 3$ .

M_{Z} (s) = e^{3 s} {(\frac{λ}{λ - 2 s})}^{3} {(\frac{λ}{λ + 4 s})}^{3}

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The pair ${X, Y}$ is iid with common moment generating function $M_{X} (s) =$ $(0.6 + 0.4 e^{s})$ . Determine the moment generating function for $Z = 5 X + 2 Y$ .

M_{Z} (s) = (0.6 + 0.4 e^{5 s}) (0.6 + 0.4 e^{2 s})

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Use the moment generating function for the symmetric triangular distribution on $(- c, c)$ as derived
in the section "Three Basic Transforms".

Obtain an expression for the symmetric triangular distribution on $(a, b)$ for any $a < b$ .
Use the result of part (a) to show that the sum of two independent random variables uniform on $(a, b)$ has symmetric triangular distribution on $(2 a, 2 b)$ .

Let $m = (a + b) / 2$ and $c = (b - a) / 2$ . If $Y \sim$ symetric triangular on $(- c, c)$ , then $X = Y + m$ is symmetric triangular on $(m - c, m + c) = (a, b)$ and

M_{X} (s) = e^{m s} M_{Y} (s) = \frac{e^{c s} + e^{- c s} - 2}{c^{2} s^{2}} e^{m s} = \frac{e^{(m + c) s} + e^{(m - c) s} - 2 e^{m s}}{c^{2} s^{2}} = \frac{e^{b s} + e^{a s} - 2 e^{\frac{a + b}{2} s}}{{(\frac{b - a}{2})}^{2} s^{2}}

M_{X + Y} (s) = {[\frac{e^{s b} - e^{s a}}{s (b - a)}]}^{2} = \frac{e^{s 2 b} + e^{s 2 a} - 2 e^{s (b + a)}}{s^{2} {(b - a)}^{2}}

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Random variable X has moment generating function $\frac{p^{2}}{{(1 - q e^{s})}^{2}}$ .

Use derivatives to determine $E [X]$ and $Var [X]$ .
Recognize the distribution from the form and compare $E [X]$ and $Var [X]$ with the result of part (a).

{[p^{2} {(1 - q e^{s})}^{- 2}]}^{'} = \frac{2 p^{2} q e^{s}}{{(1 - q e^{s})}^{3}} so that E [X] = 2 q / p

{[p^{2} {(1 - q e^{s})}^{- 2}]}^{''} = \frac{6 p^{2} q^{2} e^{s}}{{(1 - q e^{s})}^{4}} + \frac{2 p^{2} q e^{s}}{{(1 - q e^{s})}^{3}} so that E [X^{2}] = \frac{6 q^{2}}{p^{2}} + \frac{2 q}{p}

Var [X] = \frac{2 q^{2}}{p^{2}} + \frac{2 q}{p} = \frac{2 (q^{2} + p q)}{p^{2}} = \frac{2 q}{p^{2}}

$X \sim$ negative binomial $(2, p)$ , which has $E [X] = 2 q / p$ and $Var [X] = 2 q / p^{2}$ .

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The pair ${X, Y}$ is independent. $X \sim$ Poisson (4) and $Y \sim$ geometric (0.3). Determine the generating function g _Z for $Z = 3 X + 2 Y$ .

g_{Z} (s) = g_{X} (s^{3}) g_{Y} (s^{2}) = e^{4 (s^{3} - 1)} \cdot \frac{0.3}{1 - q s^{2}}

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Random variable X has moment generating function

M_{X} (s) = \frac{1}{1 - 3 s} \cdot exp (16 s^{2} / 2 + 3 s)

By recognizing forms and using rules of combinations, determine $E [X]$ and $Var [X]$ .

X = X_{1} + X_{2} with X_{1} \sim exponential(1/3) X_{2} \sim N (3, 16)

E [X] = 3 + 3 = 6 Var [X] = 9 + 16 = 25

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Source: OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6

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