(New page: 2.6 Random Sum Example. Addition of multiple independent Exponential random variables <math>\mathbf{X}_{1},\mathbf{X}_{2},\cdots</math> are <math>i.i.d.</math> Exponentail random varia...) |
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| − | 2.6 Random Sum | + | =2.6 Random Sum= |
Example. Addition of multiple independent Exponential random variables | Example. Addition of multiple independent Exponential random variables | ||
| − | <math>\mathbf{X}_{1},\mathbf{X}_{2},\cdots</math> are <math>i.i.d.</math> Exponentail random variables with parameter <math>\lambda</math> and <math>\mathbf{N}</math> is Geometric random variable with parameter <math>p</math> . Find the distribution of <math>\mathbf{S}_{\mathbf{N}}=\sum_{i=1}^{\mathbf{N}}\mathbf{X}_{i}</math> . | + | <math class="inline">\mathbf{X}_{1},\mathbf{X}_{2},\cdots</math> are <math class="inline">i.i.d.</math> Exponentail random variables with parameter <math class="inline">\lambda</math> and <math class="inline">\mathbf{N}</math> is Geometric random variable with parameter <math class="inline">p</math> . Find the distribution of <math class="inline">\mathbf{S}_{\mathbf{N}}=\sum_{i=1}^{\mathbf{N}}\mathbf{X}_{i}</math> . |
| − | Solution | + | '''Solution''' |
The probability generating function of Geometric random variable is | The probability generating function of Geometric random variable is | ||
| − | <math>P_{\mathbf{N}}\left(z\right)=E\left[z^{\mathbf{N}}\right]=\sum_{k=1}^{\infty}z^{k}\cdot q^{k-1}p=\frac{p}{q}\sum_{k=1}^{\infty}\left(z\cdot q\right)^{k}=\frac{p}{q}\cdot\frac{z\cdot q}{1-z\cdot q}=\frac{z\cdot p}{1-\cdot z\cdot q}</math>. | + | <math class="inline">P_{\mathbf{N}}\left(z\right)=E\left[z^{\mathbf{N}}\right]=\sum_{k=1}^{\infty}z^{k}\cdot q^{k-1}p=\frac{p}{q}\sum_{k=1}^{\infty}\left(z\cdot q\right)^{k}=\frac{p}{q}\cdot\frac{z\cdot q}{1-z\cdot q}=\frac{z\cdot p}{1-\cdot z\cdot q}</math>. |
The moment generating function of Exponentail random variable is | The moment generating function of Exponentail random variable is | ||
| − | <math>\phi_{\mathbf{X}}\left(s\right)=E\left[e^{s\mathbf{X}}\right]=\int_{0}^{\infty}e^{sx}\cdot\lambda e^{-\lambda x}dx=\lambda\int_{0}^{\infty}e^{x\left(s-\lambda\right)}dx=\frac{\lambda}{s-\lambda}e^{x\left(s-\lambda\right)}\biggl|_{0}^{\infty}=\frac{\lambda}{\lambda-s}.</math> | + | <math class="inline">\phi_{\mathbf{X}}\left(s\right)=E\left[e^{s\mathbf{X}}\right]=\int_{0}^{\infty}e^{sx}\cdot\lambda e^{-\lambda x}dx=\lambda\int_{0}^{\infty}e^{x\left(s-\lambda\right)}dx=\frac{\lambda}{s-\lambda}e^{x\left(s-\lambda\right)}\biggl|_{0}^{\infty}=\frac{\lambda}{\lambda-s}.</math> |
Now we can get, | Now we can get, | ||
| − | <math>\phi_{\mathbf{S}_{\mathbf{N}}}\left(s\right)=\frac{z\cdot p}{1-\cdot z\cdot q}\biggl|_{z=\frac{\lambda}{\lambda-s}}=\frac{\frac{\lambda}{\lambda-s}\cdot p}{1-\frac{\lambda}{\lambda-s}\cdot q}=\frac{\lambda p}{\lambda-s-\lambda\left(1-p\right)}=\frac{\lambda p}{\lambda p-s}.</math> | + | <math class="inline">\phi_{\mathbf{S}_{\mathbf{N}}}\left(s\right)=\frac{z\cdot p}{1-\cdot z\cdot q}\biggl|_{z=\frac{\lambda}{\lambda-s}}=\frac{\frac{\lambda}{\lambda-s}\cdot p}{1-\frac{\lambda}{\lambda-s}\cdot q}=\frac{\lambda p}{\lambda-s-\lambda\left(1-p\right)}=\frac{\lambda p}{\lambda p-s}.</math> |
| − | This is the moment generating function of Exponential random variable with parameter <math>\lambda p . Thus, f_{\mathbf{S}_{N}}\left(t\right)=p\lambda e^{-p\lambda t}.</math> | + | This is the moment generating function of Exponential random variable with parameter <math class="inline">\lambda p</math> . Thus, <math class="inline">f_{\mathbf{S}_{N}}\left(t\right)=p\lambda e^{-p\lambda t}.</math> |
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| + | ---- | ||
| + | [[ECE600|Back to ECE600]] | ||
| + | |||
| + | [[ECE 600 Sequences of Random Variables|Back to Sequences of Random Variables]] | ||
Latest revision as of 10:44, 30 November 2010
2.6 Random Sum
Example. Addition of multiple independent Exponential random variables
$ \mathbf{X}_{1},\mathbf{X}_{2},\cdots $ are $ i.i.d. $ Exponentail random variables with parameter $ \lambda $ and $ \mathbf{N} $ is Geometric random variable with parameter $ p $ . Find the distribution of $ \mathbf{S}_{\mathbf{N}}=\sum_{i=1}^{\mathbf{N}}\mathbf{X}_{i} $ .
Solution
The probability generating function of Geometric random variable is
$ P_{\mathbf{N}}\left(z\right)=E\left[z^{\mathbf{N}}\right]=\sum_{k=1}^{\infty}z^{k}\cdot q^{k-1}p=\frac{p}{q}\sum_{k=1}^{\infty}\left(z\cdot q\right)^{k}=\frac{p}{q}\cdot\frac{z\cdot q}{1-z\cdot q}=\frac{z\cdot p}{1-\cdot z\cdot q} $.
The moment generating function of Exponentail random variable is
$ \phi_{\mathbf{X}}\left(s\right)=E\left[e^{s\mathbf{X}}\right]=\int_{0}^{\infty}e^{sx}\cdot\lambda e^{-\lambda x}dx=\lambda\int_{0}^{\infty}e^{x\left(s-\lambda\right)}dx=\frac{\lambda}{s-\lambda}e^{x\left(s-\lambda\right)}\biggl|_{0}^{\infty}=\frac{\lambda}{\lambda-s}. $
Now we can get,
$ \phi_{\mathbf{S}_{\mathbf{N}}}\left(s\right)=\frac{z\cdot p}{1-\cdot z\cdot q}\biggl|_{z=\frac{\lambda}{\lambda-s}}=\frac{\frac{\lambda}{\lambda-s}\cdot p}{1-\frac{\lambda}{\lambda-s}\cdot q}=\frac{\lambda p}{\lambda-s-\lambda\left(1-p\right)}=\frac{\lambda p}{\lambda p-s}. $
This is the moment generating function of Exponential random variable with parameter $ \lambda p $ . Thus, $ f_{\mathbf{S}_{N}}\left(t\right)=p\lambda e^{-p\lambda t}. $
