/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 Let \(\\{N(t), t \geqslant 0\\}\... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 46

Let \(\\{N(t), t \geqslant 0\\}\) be a Poisson process with rate \(\lambda\), that is independent of the sequence \(X_{1}, X_{2}, \ldots\) of independent and identically distributed random variables with mean \(\mu\) and variance \(\sigma^{2}\). Find $$ \operatorname{Cov}\left(N(t), \sum_{i=1}^{N(t)} X_{i}\right) $$

Short Answer

Expert verified
The covariance between \(N(t)\) and the sum of \(N(t)\) independent and identically distributed random variables \(X_i\) is given by \(\mu\lambda t\).

Step by step solution

01

First, recall the definition of covariance: $$ \operatorname{Cov}(X,Y) = E[XY] - E[X]E[Y]. $$ Also, recall the definition of conditional expectation: $$ E[X|Y] = \sum_{x,y} x P(X=x|Y=y). $$ #Step 2: Define random variables and their properties#

Let's denote \(S = \sum_{i=1}^{N(t)} X_{i}\), and recall that \(\\{N(t), t \geqslant 0\\}\) is a Poisson process with rate \(\lambda\) and \(X_{1}, X_{2}, \ldots\) are independent and identically distributed random variables with mean \(\mu\) and variance \(\sigma^{2}\). #Step 3: Calculate \(E[N(t)]\) and \(E[S]\)#
02

As \(N(t)\) is a Poisson process, its expected value is given by: $$ E[N(t)] = \lambda t. $$ To calculate \(E[S]\), we can use the property of linearity of expectation and the fact that the Xs are identically distributed: $$ E[S] = E\left[\sum_{i=1}^{N(t)} X_{i}\right] = \sum_{i=1}^{N(t)} E[X_{i}] = N(t)\mu. $$ #Step 4: Calculate \(E[N(t)S]\) using conditional expectation#

To find \(E[N(t)S]\), we can use conditional expectation given \(N(t) = n\): $$ E[N(t)S] = E[E[N(t)S|N(t)]] = E[N(t)E[S|N(t)]]. $$ Since \(S|_{N(t)=n} = \sum_{i=1}^{n} X_{i}\), we have: $$ E[S|N(t)] = \sum_{i=1}^{N(t)} E[X_{i}] = N(t)\mu $$ Thus, $$ E[N(t)S] = E[N(t)^2\mu] = \mu E[N(t)^2]. $$ #Step 5: Calculate \(E[N(t)^2]\)#
03

For a Poisson process: $$ \operatorname{Var}(N(t)) = E[N(t)^2] - E[N(t)]^2 = \lambda t. $$ So, $$ E[N(t)^2] = \operatorname{Var}(N(t)) + E[N(t)]^2 = \lambda t + (\lambda t)^2. $$ #Step 6: Calculate the covariance#

Now, we can compute the covariance as follows: $$ \operatorname{Cov}\left(N(t), S\right) = E[N(t)S] - E[N(t)]E[S] = \mu E[N(t)^2] - (\lambda t)(N(t)\mu) = \mu(\lambda t + (\lambda t)^2) - (\lambda t)^2\mu = \mu\lambda t. $$ Therefore, the covariance between \(N(t)\) and the sum of \(N(t)\) independent and identically distributed random variables \(X_i\) is given by \(\mu\lambda t\).

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