Generating Functions for Poisson Process Distributions

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Section 8.5 Generating Functions for Poisson Process Distributions

Moment Generating Functions 5.5.1 can be derived for each of the distributions in this chapter.

Theorem 8.5.1. Moment Generating Function for Poisson.

Presuming

t > 0

and

M (t) = e^{μ (e^{t} - 1)}

Proof.

Using the Poisson probability function

\begin{aligned} M (t) & = \sum_{x = 0}^{\infty} e^{t x} \frac{μ^{x} e^{- μ}}{x!} \\ = \sum_{x = 0}^{\infty} \frac{{(μ e^{t})}^{x} e^{- μ e^{t}} e^{μ e^{t}} e^{- μ}}{x!} \\ = e^{μ e^{t}} e^{- μ} \sum_{x = 0}^{\infty} \frac{{(μ e^{t})}^{x} e^{- μ e^{t}}}{x!} \\ = e^{μ (e^{t} - 1)} \sum_{x = 0}^{\infty} \frac{{(μ e^{t})}^{x} e^{- μ e^{t}}}{x!} \\ = e^{μ (e^{t} - 1)}, \end{aligned}

where we used a new poisson distribution with new mean

μ e^{t}

to convert the sum.

Corollary 8.5.2. Poisson Properties via Moment Generating Function.

For the Poisson variable X,

M (0) = 1

M^{'} (0) = μ

M^{″} (0) = μ + μ^{2} = σ^{2} + μ^{2}

Proof.

M (0) = e^{μ (e^{0} - 1)} = e^{0} = 1.

Continuing,

M^{'} (t) = μ e^{(μ (e^{t} - 1) + t)}

and therefore

M^{'} (0) = μ e^{(μ (1 - 1) + 0)} = μ e^{0} = μ .

Continuing with the second derivative,

M^{″} (t) = (μ e^{t} + 1) μ e^{(μ (e^{t} - 1) + t)}

and therefore

M^{″} (0) = (μ + 1) μ e^{(μ (1 - 1) + 0)} = (μ + 1) μ e^{0} = μ + μ^{2}

which is the squared mean plus the variance for the poisson distribution.

Theorem 8.5.3. Moment Generating Function for Exponential.

Presuming

t < \frac{1}{μ},

M (t) = \frac{1}{1 - μ t} .

Proof.

Using the Exponential probability function 8.3.4

\begin{aligned} M (t) & = \int_{0}^{\infty} e^{t x} \frac{1}{μ} e^{- \frac{x}{μ}} d x \\ = \frac{1}{μ} \int_{0}^{\infty} e^{- x (- t + \frac{1}{μ})} d x \\ = \frac{1}{μ (- t + \frac{1}{μ})} e^{- x (- t + \frac{1}{μ})} |_{0}^{\infty} \\ = \frac{1}{μ (- t + \frac{1}{μ})} (- 0 + 1) \\ = \frac{1}{(- μ t + 1)} . \end{aligned}

Corollary 8.5.4. Exponential Properties via Moment Generating Function.

For the Exponential variable X,

M (0) = 1

M^{'} (0) = μ

M^{″} (0) = μ^{2} + μ^{2} = σ^{2} + μ^{2}

Proof.

M (0) = \frac{1}{1 - μ 0} = 1.

Continuing,

M^{'} (t) = \frac{μ}{{(1 - μ t)}^{2}}

and therefore

M^{'} (0) = \frac{μ}{{(1 - μ 0)}^{2}} = μ .

Continuing with the second derivative,

M^{″} (t) = \frac{2 μ^{2}}{{(1 - μ t)}^{3}}

and therefore

M^{″} (0) = \frac{2 μ^{2}}{{(1 - μ 0)}^{3}} = 2 μ^{2} = μ^{2} + μ^{2} .

which is the squared mean plus the variance for the poisson distribution.

Theorem 8.5.5. Moment Generating Function for Gamma.

Presuming

t < \frac{1}{μ}

where

μ

is the mean waiting time till the first "change" and

r

is the number of changes desired,

M (t) = \frac{1}{{(1 - μ t)}^{r}}

Proof.

Using the Gamma probability function 8.4.3,

\begin{aligned} M (t) & = \int_{0}^{\infty} e^{t x} \frac{x^{r - 1} \cdot e^{- \frac{x}{μ}}}{Γ (r) \cdot μ^{r}} d x \\ M (t) & = \int_{0}^{\infty} \frac{x^{r - 1} \cdot e^{- x (\frac{1}{μ} - t)}}{Γ (r) μ^{r} (\frac{1}{μ} - t)} d x \\ M (t) & = \frac{1}{{(1 - t μ)}^{r}} \int_{0}^{\infty} \frac{x^{r - 1} \cdot e^{- \frac{x}{(\frac{μ}{1 - t μ})}}}{Γ (r) \cdot {(\frac{μ}{1 - t μ})}^{r}} d x \\ = \frac{1}{{(- μ t + 1)}^{r}} . \end{aligned}

since the last integral is on the Gamma probability function but with an adjusted mean.

Corollary 8.5.6. Gamma Properties via Moment Generating Function.

For the Gamma variable X,

M (0) = 1

M^{'} (0) = r μ

M^{″} (0) = r μ^{2} + {(r μ)}^{2} = σ^{2} + μ^{2}

Proof.

M (0) = \frac{1}{{(1 - μ 0)}^{r}} \frac{1}{1} = 1.

Continuing,

M^{'} (t) = \frac{r μ}{{(1 - μ t)}^{r + 1}}

and therefore

M^{'} (0) = \frac{r μ}{{(1 - μ 0)}^{r + 1}} = r μ .

Continuing with the second derivative,

M^{″} (t) = \frac{r (r + 1) μ^{2}}{{(1 - μ t)}^{r + 2}}

and therefore

M^{″} (0) = \frac{r (r + 1) μ^{2}}{{(1 - μ 0)}^{r + 2}} = r (r + 1) μ^{2} = r μ^{2} + r^{2} μ^{2}

which is the squared mean plus the variance for the poisson distribution.

Once again, Sage can obtain the final answers quickly. For Poisson:


    
        
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var('t,mu')
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M = e^(mu *  (e^t - 1))
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Mt = derivative(M,t)
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Mtt = derivative(Mt,t)
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show(M)
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show(Mt)
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show(Mt(t=0))
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show(Mtt)
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show(Mtt(t=0))

    
    
    
    
        
            
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