13.3 Problems on transform methods (Page 2/2)

Applied probability Page 2 / 2

Random variable X has moment generating function

M_{X} (s) = \frac{exp (3 (e^{s} - 1))}{1 - 5 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} + X_{3}, with X_{1} \sim Poisson (3), X_{2} \sim exponential (1/5), X_{3} \sim N (3, 16)

E [X] = 3 + 5 + 3 = 11 Var [X] = 3 + 25 + 16 = 44

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Suppose the class ${A, B, C}$ of events is independent, with respective probabilities 0.3, 0.5, 0.2. Consider

X = - 3 I_{A} + 2 I_{B} + 4 I_{C}

Determine the moment generating functions for $I_{A}, I_{B}, I_{C}$ and use properties of moment generating functions to determine the moment generating function for X .
Use the moment generating function to determine the distribution for X .
Use canonic to determine the distribution. Compare with result (b).
Use distributions for the separate terms; determine the distribution for the sum with mgsum3. Compare with result (b).

M_{X} (s) = (0.7 + 0.3 e^{- 3 s}) (0.5 + 0.5 e^{2 s}) (0.8 + 0.2 e^{4 s}) =

0.12 e^{- 3 s} + 0.12 e^{- s} + 0.28 + 0.03 e^{s} + 0.28 e^{2 s} + 0.03 e^{3 s} + 0.07 e^{4 s} + 0.07 e^{6 s}

The distribution is

X = [- 3 - 1 0 1 2 3 4 6] P X = [0.12 0.12 0.28 0.03 0.28 0.03 0.07 0.07]

c = [-3 2 4 0];P = 0.1*[3 5 2];canonic
 Enter row vector of coefficients  cEnter row vector of minterm probabilities  minprob(P)
Use row matrices X and PX for calculationsCall for XDBN to view the distribution
P1 = [0.7 0.3];
P2 = [0.5 0.5];
P3 = [0.8 0.2];
X1 = [0 -3];
X2 = [0 2];
X3 = [0 4];
[x,px]= mgsum3(X1,X2,X3,P1,P2,P3);
disp([X;PX;x;px]')
   -3.0000    0.1200   -3.0000    0.1200-1.0000    0.1200   -1.0000    0.1200
         0    0.2800         0    0.28001.0000    0.0300    1.0000    0.0300
    2.0000    0.2800    2.0000    0.28003.0000    0.0300    3.0000    0.0300
    4.0000    0.0700    4.0000    0.07006.0000    0.0700    6.0000    0.0700

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Suppose the pair ${X, Y}$ is independent, with both X and Y binomial. Use generating functions to show under what condition, if any, $X + Y$ is binomial.

Binomial iff both have same p , as shown below.

g_{X + Y} (s) = {(q_{1} + p_{1} s)}^{n} {(q_{2} + p_{2} s)}^{m} = {(q + p s)}^{n + m} iff p_{1} = p_{2}

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Suppose the pair ${X, Y}$ is independent, with both X and Y Poisson.

Use generating functions to show under what condition $X + Y$ is Poisson.
What about $X - Y$ ? Justify your answer.

Always Poisson, as the argument below shows.

g_{X + Y} (s) = e^{μ (s - 1)} e^{ν (s - 1)} = e^{(μ + ν) (s - 1)}

However, $Y - X$ could have negative values.

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Suppose the pair ${X, Y}$ is independent, Y is nonnegative integer-valued, X is Poisson and $X + Y$ is Poisson. Use the generating functions to show that Y is Poisson.

$E [X + Y] = μ + ν$ , where $ν = E [Y] > 0$ . $g_{X} (s) = e^{μ (s - 1)}$ and $g_{X + Y} (s) = g_{X} (s) g_{Y} (s) = e^{(μ + ν) (s - 1)}$ . Division by $g_{X} (s)$ gives $g_{Y} (s) = e^{ν (s - 1)}$ .

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Suppose the pair ${X, Y}$ is iid, binomial $(6, 0.51)$ . By the result of [link]

$X + Y$ is binomial. Use mgsum to obtain the distribution for $Z = 2 X + 4 Y$ . Does Z have the binomial distribution? Is the result surprising? Examine the first few possible values for Z . Write the generating function for Z ; does it have the form for the binomial distribution?

x  = 0:6;
px = ibinom(6,0.51,x);[Z,PZ] = mgsum(2*x,4*x,px,px);disp([Z(1:5);PZ(1:5)]')0    0.0002       % Cannot be binomial, since odd values missing
    2.0000    0.00124.0000    0.0043
    6.0000    0.01188.0000    0.0259
    - - - - - - - -

g_{X} (s) = g_{Y} (s) = {(0.49 + 0.51 s)}^{6} g_{Z} (s) = {(0.49 + 0.51 s^{2})}^{6} {(0.49 + 0.51 s^{4})}^{6}

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Suppose the pair ${X, Y}$ is independent, with $X \sim$ binomial $(5, 0.33)$ and

$Y \sim$ binomial $(7, 0.47)$ .

Let $G = g (X) = 3 X^{2} - 2 X$ and $H = h (Y) = 2 Y^{2} + Y + 3$ .

Use the mgsum to obtain the distribution for $G + H$ .
Use icalc and csort to obtain the distribution for $G + H$ and compare with the result of part (a).

X = 0:5;
Y = 0:7;PX = ibinom(5,0.33,X);
PY = ibinom(7,0.47,Y);G = 3*X.^2 - 2*X;
H = 2*Y.^2 + Y + 3;[Z,PZ] = mgsum(G,H,PX,PY);icalc
Enter row matrix of X-values  XEnter row matrix of Y-values  Y
Enter X probabilities  PXEnter Y probabilities  PY
 Use array operations on matrices X, Y, PX, PY, t, u, and PM = 3*t.^2 - 2*t + 2*u.^2 + u + 3;
[z,pz]= csort(M,P);
e = max(abs(pz - PZ))  % Comparison of p valuese =  0

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Suppose the pair ${X, Y}$ is independent, with $X \sim$ binomial $(8, 0.39)$ and

$Y \sim$ uniform on ${- 1.3, - 0.5, 1.3, 2.2, 3.5}$ . Let

U = 3 X^{2} - 2 X + 1 and V = Y^{3} + 2 Y - 3

Use mgsum to obtain the distribution for $U + V$ .
Use icalc and csort to obtain the distribution for $U + V$ and compare with the result of part (a).

X = 0:8;
Y = [-1.3 -0.5 1.3 2.2 3.5];
PX = ibinom(8,0.39,X);PY = (1/5)*ones(1,5);
U  = 3*X.^2 - 2*X + 1;V  = Y.^3 + 2*Y - 3;
[Z,PZ]= mgsum(U,V,PX,PY);
icalcEnter row matrix of X-values  X
Enter row matrix of Y-values  YEnter X probabilities  PX
Enter Y probabilities  PYUse array operations on matrices X, Y, PX, PY, t, u, and P
M = 3*t.^2 - 2*t + 1 + u.^3 + 2*u - 3;[z,pz] = csort(M,P);e = max(abs(pz - PZ))
e = 0

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If X is a nonnegative integer-valued random variable, express the generating function as a power series.

Show that the k th derivative at $s = 1$ is
$g_{X}^{(k)} (1) = E [X (X - 1) (X - 2) \dots (X - k + 1)]$
Use this to show the $Var [X] = g_{X}^{''} (1) + g_{X}^{'} (1) - {[g_{X}^{'} (1)]}^{2}$ .

Since power series may be differentiated term by term

g_{X}^{(n)} (s) = \sum_{k = n}^{\infty} k (k - 1) \dots (k - n + 1) p_{k} s^{k - n} so that

g_{X}^{(n)} (1) = \sum_{k = n}^{\infty} k (k - 1) \dots (k - n + 1) p_{k} = E [X (X - 1) \dots (X - n + 1)]

Var [X] = E [X^{2}] - E^{2} [X] = E [X (X - 1)] + E [X] - E^{2} [X] = g_{X}^{''} (1) + g_{X}^{'} (1) - {[g_{X}^{'} (1)]}^{2}

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Let $M_{X} (\cdot)$ be the moment generating function for X .

Show that $Var [X]$ is the second derivative of $e^{- s μ} M_{X} (s)$ evaluated at $s = 0$ .
Use this fact to show that if $X \sim N (μ, σ^{2})$ , then $Var [X] = σ^{2}$ .

f (s) = e^{- s μ} M_{X} (s) f^{''} (s) = e^{- s μ} [- μ M_{X}^{'} (s) + μ^{2} M_{X} (s) + M_{X}^{''} (s) - μ M_{X}^{'} (s)]

Setting $s = 0$ and using the result on moments gives

f^{''} (0) = - μ^{2} + μ^{2} + E [X^{2}] - μ^{2} = Var [X]

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Use derivatives of $M_{X_{m}} (s)$ to obtain the mean and variance of the negative binomial $(m, p)$ distribution.

To simplify writing use $f (s)$ for $M_{X} (S)$ .

f (s) = \frac{p^{m}}{{(1 - q e^{s})}^{m}} f^{'} (s) = \frac{m p^{m} q e^{s}}{{(1 - q e^{s})}^{m + 1}} f^{''} (s) = \frac{m p^{m} q e^{s}}{{(1 - q e^{s})}^{m + 1}} + \frac{m (m + 1) p^{m} q^{2} e^{2 s}}{{(1 - q e^{s})}^{m + 2}}

E [X] = \frac{m p^{m} q}{{(1 - q)}^{m + 1}} = \frac{m q}{p} E [X^{2}] = \frac{m q}{p} + \frac{m (m + 1) p^{m} q^{2}}{{(1 - q)}^{m + 2}}

Var [X] = \frac{m q}{p} + \frac{m (m + 1) q^{2}}{p^{2}} - \frac{m^{2} q^{2}}{p^{2}} = \frac{m q}{p^{2}}

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Use moment generating functions to show that variances add for the sum or difference of independent random variables.

To simplify writing, set $f (s) = M_{X} (s)$ , $g (s) = M_{Y} (s)$ , and $h (s) = M_{X} (s) M_{Y} (s)$

h^{'} (s) = f^{'} (s) g (s) + f (s) g^{'} (s) h^{''} (s) = f^{''} (s) g (s) + f^{'} (s) g^{'} (s) + f^{'} (s) g^{'} (s) + f (s) g^{''} (s)

Setting $s = 0$ yields

E [X + Y] = E [X] + E [Y] E [{(X + Y)}^{2}] = E [X^{2}] + 2 E [X] E [Y] + E [Y^{2}] E^{2} [X + Y] =

E^{2} [X] + 2 E [X] E [Y] + E^{2} [Y]

Taking the difference gives $Var [X + Y] = Var [X] + Var [Y]$ . A similar treatment with $g (s)$ replaced by $g (- s)$ shows $Var [X - Y] = Var [X] + Var [Y]$ .

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The pair ${X, Y}$ is iid $N (3, 5)$ . Use the moment generating function to show that $Z = 3 X - 2 Y + 3$ is is normal (see Example 3 from "Transform Methods" for general result).

M_{3 X} (s) = M_{X} (3 s) = exp (\frac{9 \cdot 5 s^{2}}{2} + 3 \cdot 3 s) M_{- 2 Y} (s) = M_{Y} (- 2 s) = exp (\frac{4 \cdot 5 s^{2}}{2} - 2 \cdot 3 s)

M_{Z} (s) = e^{3 s} exp (\frac{(45 + 20) s^{2}}{2} + (9 - 6) s) = exp (\frac{65 s^{2}}{2} + 6 s)

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Use the central limit theorem to show that for large enough sample size (usually 20 or more), the sample average

A_{n} = \frac{1}{n} \sum_{i = 1}^{n} X_{i}

is approximately $N (μ, σ^{2} / n)$ for any reasonable population distribution having mean value μ and variance σ ² .

E [A_{n}] = \frac{1}{n} \sum_{i = 1}^{n} μ = μ Var [A_{n}] = \frac{1}{n^{2}} \sum_{i = 1}^{n} σ^{2} = \frac{σ^{2}}{n}

By the central limit theorem, A _n is approximately normal, with the mean and variance above.

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A population has standard deviation approximately three. It is desired to determine the sample size n needed to ensure that with probability 0.95 the sample average will be within 0.5 of the mean value.

Use the Chebyshev inequality to estimate the needed sample size.
Use the normal approximation to estimate n (see Example 1 from "Simple Random Samples and Statistics").

Chebyshev inequality
$P (\frac{| A_{n} - μ |}{σ / \sqrt{n}} \geq \frac{0.5 \sqrt{n}}{3}) \leq \frac{3^{2}}{0 . 5^{2} n} \leq 0.05 implies n \geq 720$
Normal approximation Use of the table in Example 1 from "Simple Random Samples and Statistics" shows
$n \geq {(3 / 0.5)}^{2} 3.84 = 128$

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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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