Revision as of 12:12, 23 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] $ [CS1GeometricDistribution].

(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}}. $

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