(New page: 2.4 Strong law of large numbers (Borel) Let <math>\left\{ \mathbf{X}_{n}\right\}</math> be a sequence of identically distributed random variables with mean <math>\mu</math> and varianc...) |
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2.4 Strong law of large numbers (Borel) | 2.4 Strong law of large numbers (Borel) | ||
| − | Let <math>\left\{ \mathbf{X}_{n}\right\}</math> be a sequence of identically distributed random variables with mean <math>\mu</math> and variance <math>\sigma^{2}</math> , and <math>Cov\left(\mathbf{X}_{i},\mathbf{X}_{j}\right)=E\left[\left(\mathbf{X}_{i}-\mu\right)\left(\mathbf{X}_{j}-\mu\right)\right]=0,\quad i\neq j\text{ : uncorrelated.}</math> | + | Let <math class="inline">\left\{ \mathbf{X}_{n}\right\}</math> be a sequence of identically distributed random variables with mean <math class="inline">\mu</math> and variance <math class="inline">\sigma^{2}</math> , and <math class="inline">Cov\left(\mathbf{X}_{i},\mathbf{X}_{j}\right)=E\left[\left(\mathbf{X}_{i}-\mu\right)\left(\mathbf{X}_{j}-\mu\right)\right]=0,\quad i\neq j\text{ : uncorrelated.}</math> |
| − | Then <math>\mathbf{Y}_{n}=\frac{1}{n}\sum_{k=1}^{n}\mathbf{X}_{k}\longrightarrow\left(a.e.\right)\longrightarrow\mu\text{ as }n\longrightarrow\infty.</math> | + | Then <math class="inline">\mathbf{Y}_{n}=\frac{1}{n}\sum_{k=1}^{n}\mathbf{X}_{k}\longrightarrow\left(a.e.\right)\longrightarrow\mu\text{ as }n\longrightarrow\infty.</math> |
Proof | Proof | ||
Beyound this course. Require measure theory. | Beyound this course. Require measure theory. | ||
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| + | ---- | ||
| + | [[ECE600|Back to ECE600]] | ||
| + | |||
| + | [[ECE 600 Sequences of Random Variables|Back to Sequences of Random Variables]] | ||
Latest revision as of 11:41, 30 November 2010
2.4 Strong law of large numbers (Borel)
Let $ \left\{ \mathbf{X}_{n}\right\} $ be a sequence of identically distributed random variables with mean $ \mu $ and variance $ \sigma^{2} $ , and $ Cov\left(\mathbf{X}_{i},\mathbf{X}_{j}\right)=E\left[\left(\mathbf{X}_{i}-\mu\right)\left(\mathbf{X}_{j}-\mu\right)\right]=0,\quad i\neq j\text{ : uncorrelated.} $
Then $ \mathbf{Y}_{n}=\frac{1}{n}\sum_{k=1}^{n}\mathbf{X}_{k}\longrightarrow\left(a.e.\right)\longrightarrow\mu\text{ as }n\longrightarrow\infty. $
Proof
Beyound this course. Require measure theory.
