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Chapter 5: Q78SE (page 244)

According to the article 鈥淩eliability Evaluation of Hard Disk Drive Failures Based on Counting Processes鈥� (Reliability Engr. and System Safety,\({\bf{2013}}:{\rm{ }}{\bf{110}}--{\bf{118}}\)), particles accumulating on a disk drive come from two sources, one external and the other internal. The article proposed a model in which the internal source contains a number of loose particles W having a Poisson distribution with mean value of m; when a loose particle releases, it immediately enters the drive, and the release times are independent and identically distributed with cumulative distribution function G(t). Let X denote the number of loose particles not yet released at a particular time t. Show that X has a Poisson distribution with parameter\(\mu \left( {1 - G\left( t \right)} \right)\). (Hint: Let Y denote the number of particles accumulated on the drive from the internal source by time t so that\(X + Y = W\). Obtain an expression for\(P\left( {X = x, Y = y} \right)\)and then sum over y.)

Short Answer

Expert verified

The expression for the following question is \(X\~{\mathop{\rm Poisson}\nolimits} (\mu (1 - G(t)))\).

Step by step solution

01

Definition of Poisson Distribution.

The Poisson distribution is a discrete probability distribution in probability theory and statistics that reflects the likelihood of a specific number of events occurring in a set interval of time or space at a known constant mean rate, regardless of the time since the last event.

02

Explanation for obtaining the expression mentioned in the question.

Given:

\(\begin{array}{l}W\sim{\mathop{\rm Poisson}\nolimits} (\mu )\\X + Y = W\end{array}\)

Formula Poisson probability:

\(P(W = w) = \frac{{{\mu ^w}{e^{ - \mu }}}}{{w!}}\)

The number of successes among a fixed number of the independent trials of fixed probability of success follows a binomial distribution. In this case, the constant probability of failure is\(G\left( t \right)\)thus the constant probability of success is\((1 - G(t))\).

\(\begin{array}{c}P(X = x,Y = y\mid W = x + y) = \left( {\begin{array}{*{20}{c}}{x + y}\\x\end{array}} \right){(1 - G(t))^x}G{(t)^{(x + y) - x}}\\ = \left( {\begin{array}{*{20}{c}}{x + y}\\x\end{array}} \right){(1 - G(t))^x}G{(t)^y}\end{array}\)

03

Explanation for obtaining the expression mentioned in the question.

Use the General multiplication rule:

\(P(A\;and\;B) = P(A) \times P(B\mid A) = P(B) \times P(A\mid B)\)

\(\begin{array}{l}P(X = x,Y = y) = P(X = x,Y = y\mid W = x + y)P(W = x + y)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \left( {\begin{array}{*{20}{c}}{x + y}\\x\end{array}} \right){(1 - G(t))^x}G{(t)^y}\frac{{{\mu ^{x + y}}{e^{ - \mu }}}}{{(x + y)!}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \frac{{(x + y)!}}{{x!y!}}{(1 - G(t))^x}G{(t)^y}\frac{{{\mu ^x}{\mu _y}{e^{ - \mu }}}}{{(x + y)!}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \frac{{{\mu ^x}{\mu _y}{e^{ - \mu }}}}{{x!y!}}{(1 - G(t))^x}G{(t)^y}\end{array}\)

04

Explanation for obtaining the expression mentioned in the question.

The probability distribution of X is then obtained by summing over all possible values of y:

\(\begin{array}{c}P(X = x) = \sum\limits_{y = 0}^{ + \infty } P (X = x,Y = y)\\ = \sum\limits_{y = 0}^{ + \infty } {\frac{{{\mu ^x}{\mu _y}{e^{ - \mu }}}}{{x!y!}}} G{(t)^x}{(1 - G(t))^y}\\ = \frac{{{\mu ^x}{{(1 - G(t))}^x}}}{{x!}}\sum\limits_{y = 0}^{ + \infty } {\frac{{{\mu ^y}{e^{ - \mu (G(t) + (1 - G(t)))}}G{{(t)}^y}}}{{y!}}} \\ = \frac{{{{(\mu (1 - G(t)))}^x}{e^{ - \mu (1 - G(t))}}}}{{x!}}\sum\limits_{y = 0}^{ + \infty } {\frac{{{{(\mu G(t))}^y}{e^{ - \mu G(t)}}}}{{y!}}} \end{array}\)

\(P(Y = y) = \frac{{{{(\mu G(t))}^y}{e^{ - \mu G(t)}}}}{{y!}}\)is pdf of\({\mathop{\rm Poisson}\nolimits} (\mu G(t))\)

Sum of all probabilities of valid pdf is 1

\(\begin{array}{l} = \frac{{{{(\mu (1 - G(t)))}^x}{e^{ - \mu (1 - G(t))}}}}{{x!}}(1)\\ = \frac{{{{(\mu (1 - G(t)))}^x}{e^{ - \mu (1 - G(t))}}}}{{x!}}\end{array}\)

\(P(X = x) = \frac{{{{(\mu (1 - G(t)))}^x}{e^{ - \mu (1 - G(t))}}}}{{x!}}\)is the probability density function of a Poisson distribution with parameter\(\mu (1 - G(t))\).

\(X\sim {\mathop{\rm Poisson}\nolimits} (\mu (1 - G(t)))\)

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Most popular questions from this chapter

A surveyor wishes to lay out a square region with each side having length\({\rm{L}}\). However, because of a measurement error, he instead lays out a rectangle in which the north-south sides both have length \({\rm{X}}\) and the east-west sides both have length\({\rm{Y}}\). Suppose that \({\rm{X}}\) and \({\rm{Y}}\) are independent and that each is uniformly distributed on the interval \({\rm{(L - A,L + A)}}\) (where \({\rm{0 < A < L}}\) ). What is the expected area of the resulting rectangle?

Let \({\rm{X}}\) denote the number of Canon SLR cameras sold during a particular week by a certain store. The pmf of \({\rm{X}}\) is

Sixty percent of all customers who purchase these cameras also buy an extended warranty. Let \({\rm{Y}}\) denote the number of purchasers during this week who buy an extended warranty.

a. What is\({\rm{P(X = 4,Y = 2)}}\)? (Hint: This probability equals\({\rm{P(Y = 2}}\mid {\rm{X = 4) \times P(X = 4)}}\); now think of the four purchases as four trials of a binomial experiment, with success on a trial corresponding to buying an extended warranty.)

b. Calculate\({\rm{P(X = Y)}}\).

c. Determine the joint pmf of \({\rm{X}}\) and \({\rm{Y}}\)then the marginal pmf of\({\rm{Y}}\).

a. Let \({{\rm{X}}_{\rm{1}}}\)have a chi-squared distribution with parameter \({{\rm{\nu }}_{\rm{1}}}\) (see Section 4.4), and let \({{\rm{X}}_{\rm{2}}}\)be independent of \({{\rm{X}}_{\rm{1}}}\)and have a chi-squared distribution with parameter\({{\rm{v}}_{\rm{2}}}\). Use the technique of to show that \({{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}\)has a chi-squared distribution with parameter\({{\rm{v}}_{\rm{1}}}{\rm{ + }}{{\rm{v}}_{\rm{2}}}\).

b. You were asked to show that if \({\rm{Z}}\)is a standard normal \({\rm{rv}}\), then \({{\rm{Z}}^{\rm{2}}}\)has a chi squared distribution with\({\rm{v = 1}}\). Let \({{\rm{Z}}_{\rm{1}}}{\rm{,}}{{\rm{Z}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{Z}}_{\rm{n}}}\)be \({\rm{n}}\)independent standard normal \({\rm{rv}}\) 's. What is the distribution of\({\rm{Z}}_{\rm{1}}^{\rm{2}}{\rm{ + }}...{\rm{ + Z}}_{\rm{n}}^{\rm{2}}\)? Justify your answer.

c. Let \({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)be a random sample from a normal distribution with mean \({\rm{\mu }}\)and variance\({{\rm{\sigma }}^{\rm{2}}}\). What is the distribution of the sum \({\rm{Y = }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {\left( {{{\rm{X}}_{\rm{i}}}{\rm{ - \mu }}} \right){\rm{/\sigma }}} \right)}^{\rm{2}}}} {\rm{?}}\)Justify your answer?

Six individuals, including \({\rm{A}}\) and \({\rm{B}}\), take seats around a circular table in a completely random fashion. Suppose the seats are numbered\({\rm{1, \ldots ,6}}\). Let \({\rm{X = }}\) A's seat number and \({\rm{Y = B}}\) 's seat number. If A sends a written message around the table to \({\rm{B}}\) in the direction in which they are closest, how many individuals (including A and B) would you expect to handle the message?

Carry out a simulation experiment using a statistical computer package or other software to study the sampling distribution of \({\rm{\bar X}}\) when the population distribution is lognormal with \({\rm{E(ln(X)) = 3}}\) and\({\rm{V(ln(X)) = 1}}\). Consider the four sample sizes\({\rm{n = 10,20,30}}\), and\({\rm{50}}\), and in each case use \({\rm{1000}}\) replications. For which of these sample sizes does the \({\rm{\bar X}}\) sampling distribution appear to be approximately normal?
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