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=Example. Geometric random variable= | =Example. Geometric random variable= | ||
| − | Let <math>\mathbf{X}</math> be a random variable with probability mass function | + | Let <math class="inline">\mathbf{X}</math> be a random variable with probability mass function |
| − | <math>p_{\mathbf{X}}\left(k\right)=\alpha\left(1-\alpha\right)^{k-1},k=1,2,3,\cdots</math> | + | <math class="inline">p_{\mathbf{X}}\left(k\right)=\alpha\left(1-\alpha\right)^{k-1},k=1,2,3,\cdots</math> |
| − | where <math>0<\alpha<1</math> . | + | where <math class="inline">0<\alpha<1</math> . |
Note | Note | ||
| − | This is a geometric random variable with success probability <math>\alpha</math> . | + | This is a geometric random variable with success probability <math class="inline">\alpha</math> . |
| − | (a) Find the characteristic function of <math>\mathbf{X}</math> . | + | (a) Find the characteristic function of <math class="inline">\mathbf{X}</math> . |
| − | <math>\Phi_{\mathbf{X}}\left(\omega\right)=E\left[e^{i\omega\mathbf{X}}\right]=\sum_{k=1}^{\infty}e^{i\omega k}\alpha\left(1-\alpha\right)^{k-1}=\alpha e^{i\omega}\sum_{k=1}^{\infty}\left[e^{i\omega}\left(1-\alpha\right)\right]^{k-1}</math><math>=\alpha e^{i\omega}\sum_{m=0}^{\infty}\left[e^{i\omega}\left(1-\alpha\right)\right]^{m}=\frac{\alpha e^{i\omega}}{1-e^{i\omega}\left(1-\alpha\right)}\text{ (infinite geometric series)}</math> | + | <math class="inline">\Phi_{\mathbf{X}}\left(\omega\right)=E\left[e^{i\omega\mathbf{X}}\right]=\sum_{k=1}^{\infty}e^{i\omega k}\alpha\left(1-\alpha\right)^{k-1}=\alpha e^{i\omega}\sum_{k=1}^{\infty}\left[e^{i\omega}\left(1-\alpha\right)\right]^{k-1}</math><math class="inline">=\alpha e^{i\omega}\sum_{m=0}^{\infty}\left[e^{i\omega}\left(1-\alpha\right)\right]^{m}=\frac{\alpha e^{i\omega}}{1-e^{i\omega}\left(1-\alpha\right)}\text{ (infinite geometric series)}</math> |
| − | since <math>\left|e^{i\omega}\left(1-\alpha\right)\right|<1</math> . | + | since <math class="inline">\left|e^{i\omega}\left(1-\alpha\right)\right|<1</math> . |
| − | <math>\because0<1-\alpha<1</math> and the real term of <math>e^{i\omega}=\cos\omega+i\sin\omega</math> is <math>\left|\cos\omega\right|<1</math> . | + | <math class="inline">\because0<1-\alpha<1</math> and the real term of <math class="inline">e^{i\omega}=\cos\omega+i\sin\omega</math> is <math class="inline">\left|\cos\omega\right|<1</math> . |
| − | (b) Find the mean of <math>\mathbf{X}</math> . | + | (b) Find the mean of <math class="inline">\mathbf{X}</math> . |
| − | <math>E\left[\mathbf{X}\right]=\frac{d}{d\left(i\omega\right)}\Phi_{\mathbf{X}}\left(\omega\right)\left|_{i\omega=0}\right.=\frac{d}{d\left(i\omega\right)}\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}\left|_{i\omega=0}\right.</math><math>=\left[\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}+\alpha e^{i\omega}\cdot\left(-1\right)\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\cdot\left(-e^{i\omega}\left(1-\alpha\right)\right)\right]\left|_{i\omega=0}\right.</math><math>=\left[\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}+\alpha\left(1-\alpha\right)e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\right]\left|_{i\omega=0}\right.</math><math>=\alpha\left(1-\left(1-\alpha\right)\right)^{-1}+\alpha\left(1-\alpha\right)\left(1-\left(1-\alpha\right)\right)^{-2}=1+\frac{\left(1-\alpha\right)}{\alpha}=\frac{1}{\alpha}.</math> | + | <math class="inline">E\left[\mathbf{X}\right]=\frac{d}{d\left(i\omega\right)}\Phi_{\mathbf{X}}\left(\omega\right)\left|_{i\omega=0}\right.=\frac{d}{d\left(i\omega\right)}\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}\left|_{i\omega=0}\right.</math><math class="inline">=\left[\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}+\alpha e^{i\omega}\cdot\left(-1\right)\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\cdot\left(-e^{i\omega}\left(1-\alpha\right)\right)\right]\left|_{i\omega=0}\right.</math><math class="inline">=\left[\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}+\alpha\left(1-\alpha\right)e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\right]\left|_{i\omega=0}\right.</math><math class="inline">=\alpha\left(1-\left(1-\alpha\right)\right)^{-1}+\alpha\left(1-\alpha\right)\left(1-\left(1-\alpha\right)\right)^{-2}=1+\frac{\left(1-\alpha\right)}{\alpha}=\frac{1}{\alpha}.</math> |
Note | Note | ||
| − | You can see the other approach to find <math>E\left[\mathbf{X}\right]</math> and <math>Var\left[\mathbf{X}\right]</math> [ | + | You can see the other approach to find <math class="inline">E\left[\mathbf{X}\right]</math> and <math class="inline">Var\left[\mathbf{X}\right]</math> [[ECE 600 Prerequisites Discrete Random Variables|(See Geometric Distribution)]]. |
| − | (c) Find the variance of <math>\mathbf{X}</math> . | + | (c) Find the variance of <math class="inline">\mathbf{X}</math> . |
| − | <math>E\left[\mathbf{X}^{2}\right]=\frac{d^{2}}{d\left(i\omega\right)^{2}}\Phi_{\mathbf{X}}\left(\omega\right)\left|_{i\omega=0}\right.=\frac{d^{2}}{d\left(i\omega\right)^{2}}\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}\left|_{i\omega=0}\right.</math><math>=\frac{d}{d\left(i\omega\right)}\left[\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}+\alpha\left(1-\alpha\right)e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\right]\left|_{i\omega=0}\right.</math><math>=\frac{d}{d\left(i\omega\right)}\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}\left|_{i\omega=0}\right.+\alpha\left(1-\alpha\right)\frac{d}{d\left(i\omega\right)}e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\left|_{i\omega=0}\right.</math><math>=\frac{1}{\alpha}+\alpha\left(1-\alpha\right)\frac{2}{\alpha^{3}}=\frac{\alpha}{\alpha^{2}}+\frac{2-2\alpha}{\alpha^{2}}=\frac{2-\alpha}{\alpha^{2}}</math> | + | <math class="inline">E\left[\mathbf{X}^{2}\right]=\frac{d^{2}}{d\left(i\omega\right)^{2}}\Phi_{\mathbf{X}}\left(\omega\right)\left|_{i\omega=0}\right.=\frac{d^{2}}{d\left(i\omega\right)^{2}}\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}\left|_{i\omega=0}\right.</math><math class="inline">=\frac{d}{d\left(i\omega\right)}\left[\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}+\alpha\left(1-\alpha\right)e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\right]\left|_{i\omega=0}\right.</math><math class="inline">=\frac{d}{d\left(i\omega\right)}\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}\left|_{i\omega=0}\right.+\alpha\left(1-\alpha\right)\frac{d}{d\left(i\omega\right)}e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\left|_{i\omega=0}\right.</math><math class="inline">=\frac{1}{\alpha}+\alpha\left(1-\alpha\right)\frac{2}{\alpha^{3}}=\frac{\alpha}{\alpha^{2}}+\frac{2-2\alpha}{\alpha^{2}}=\frac{2-\alpha}{\alpha^{2}}</math> |
because | because | ||
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| − | <math>\frac{d}{d\left(i\omega\right)}e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\left|_{i\omega=0}\right. =2e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}+e^{i\omega2}\left(-2\right)\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-3}\left(-e^{i\omega}\left(1-\alpha\right)\right)\left|_{i\omega=0}\right.</math><math>=2e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}+2\left(1-\alpha\right)e^{i\omega3}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-3}\left|_{i\omega=0}\right.</math><math>=2\alpha^{-2}+2\left(1-\alpha\right)\alpha^{-3}=\frac{2\alpha+2-2\alpha}{\alpha^{3}}=\frac{2}{\alpha^{3}}.</math> | + | <math class="inline">\frac{d}{d\left(i\omega\right)}e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\left|_{i\omega=0}\right. =2e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}+e^{i\omega2}\left(-2\right)\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-3}\left(-e^{i\omega}\left(1-\alpha\right)\right)\left|_{i\omega=0}\right.</math><math class="inline">=2e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}+2\left(1-\alpha\right)e^{i\omega3}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-3}\left|_{i\omega=0}\right.</math><math class="inline">=2\alpha^{-2}+2\left(1-\alpha\right)\alpha^{-3}=\frac{2\alpha+2-2\alpha}{\alpha^{3}}=\frac{2}{\alpha^{3}}.</math> |
| − | <math>Var\left[\mathbf{X}\right]=E\left[\mathbf{X}^{2}\right]-\left(E\left[\mathbf{X}\right]\right)^{2}=\frac{2-\alpha}{\alpha^{2}}-\frac{1}{\alpha^{2}}=\frac{1-\alpha}{\alpha^{2}}.</math> | + | <math class="inline">Var\left[\mathbf{X}\right]=E\left[\mathbf{X}^{2}\right]-\left(E\left[\mathbf{X}\right]\right)^{2}=\frac{2-\alpha}{\alpha^{2}}-\frac{1}{\alpha^{2}}=\frac{1-\alpha}{\alpha^{2}}.</math> |
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Latest revision as of 11:10, 30 November 2010
Example. Geometric random variable
Let $ \mathbf{X} $ be a random variable with probability mass function
$ p_{\mathbf{X}}\left(k\right)=\alpha\left(1-\alpha\right)^{k-1},k=1,2,3,\cdots $
where $ 0<\alpha<1 $ .
Note
This is a geometric random variable with success probability $ \alpha $ .
(a) Find the characteristic function of $ \mathbf{X} $ .
$ \Phi_{\mathbf{X}}\left(\omega\right)=E\left[e^{i\omega\mathbf{X}}\right]=\sum_{k=1}^{\infty}e^{i\omega k}\alpha\left(1-\alpha\right)^{k-1}=\alpha e^{i\omega}\sum_{k=1}^{\infty}\left[e^{i\omega}\left(1-\alpha\right)\right]^{k-1} $$ =\alpha e^{i\omega}\sum_{m=0}^{\infty}\left[e^{i\omega}\left(1-\alpha\right)\right]^{m}=\frac{\alpha e^{i\omega}}{1-e^{i\omega}\left(1-\alpha\right)}\text{ (infinite geometric series)} $
since $ \left|e^{i\omega}\left(1-\alpha\right)\right|<1 $ .
$ \because0<1-\alpha<1 $ and the real term of $ e^{i\omega}=\cos\omega+i\sin\omega $ is $ \left|\cos\omega\right|<1 $ .
(b) Find the mean of $ \mathbf{X} $ .
$ E\left[\mathbf{X}\right]=\frac{d}{d\left(i\omega\right)}\Phi_{\mathbf{X}}\left(\omega\right)\left|_{i\omega=0}\right.=\frac{d}{d\left(i\omega\right)}\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}\left|_{i\omega=0}\right. $$ =\left[\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}+\alpha e^{i\omega}\cdot\left(-1\right)\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\cdot\left(-e^{i\omega}\left(1-\alpha\right)\right)\right]\left|_{i\omega=0}\right. $$ =\left[\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}+\alpha\left(1-\alpha\right)e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\right]\left|_{i\omega=0}\right. $$ =\alpha\left(1-\left(1-\alpha\right)\right)^{-1}+\alpha\left(1-\alpha\right)\left(1-\left(1-\alpha\right)\right)^{-2}=1+\frac{\left(1-\alpha\right)}{\alpha}=\frac{1}{\alpha}. $
Note
You can see the other approach to find $ E\left[\mathbf{X}\right] $ and $ Var\left[\mathbf{X}\right] $ (See Geometric Distribution).
(c) Find the variance of $ \mathbf{X} $ .
$ E\left[\mathbf{X}^{2}\right]=\frac{d^{2}}{d\left(i\omega\right)^{2}}\Phi_{\mathbf{X}}\left(\omega\right)\left|_{i\omega=0}\right.=\frac{d^{2}}{d\left(i\omega\right)^{2}}\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}\left|_{i\omega=0}\right. $$ =\frac{d}{d\left(i\omega\right)}\left[\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}+\alpha\left(1-\alpha\right)e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\right]\left|_{i\omega=0}\right. $$ =\frac{d}{d\left(i\omega\right)}\alpha e^{i\omega}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-1}\left|_{i\omega=0}\right.+\alpha\left(1-\alpha\right)\frac{d}{d\left(i\omega\right)}e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\left|_{i\omega=0}\right. $$ =\frac{1}{\alpha}+\alpha\left(1-\alpha\right)\frac{2}{\alpha^{3}}=\frac{\alpha}{\alpha^{2}}+\frac{2-2\alpha}{\alpha^{2}}=\frac{2-\alpha}{\alpha^{2}} $
because
$ \frac{d}{d\left(i\omega\right)}e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}\left|_{i\omega=0}\right. =2e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}+e^{i\omega2}\left(-2\right)\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-3}\left(-e^{i\omega}\left(1-\alpha\right)\right)\left|_{i\omega=0}\right. $$ =2e^{i\omega2}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-2}+2\left(1-\alpha\right)e^{i\omega3}\left(1-e^{i\omega}\left(1-\alpha\right)\right)^{-3}\left|_{i\omega=0}\right. $$ =2\alpha^{-2}+2\left(1-\alpha\right)\alpha^{-3}=\frac{2\alpha+2-2\alpha}{\alpha^{3}}=\frac{2}{\alpha^{3}}. $
$ Var\left[\mathbf{X}\right]=E\left[\mathbf{X}^{2}\right]-\left(E\left[\mathbf{X}\right]\right)^{2}=\frac{2-\alpha}{\alpha^{2}}-\frac{1}{\alpha^{2}}=\frac{1-\alpha}{\alpha^{2}}. $
