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Exercise 5.3.8

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When $t \leq x_{1}$ , $P (x > t) = 1$ , if $x_{1} < t < x_{2}$ , $P (x > t) = 1 - p_{1}$ , if $x_{i} < t < x_{i + 1}$ , then $P (x > t) = 1 - \sum_{k = 1}^{i} p_{k}$

\begin{array}{l} \int_{t = 0}^{\infty} P (x > t) dt & = \int_{t = 0}^{x_{1}} P (x > t) dt + \int_{x_{1}}^{x_{2}} P (x > t) dt + \dots + \int_{x_{i}}^{x_{i + 1}} P (x > t) dt + \dots + \int_{x_{n}}^{\infty} P (x > t) dt \\ = 1 x_{1} + (1 - p_{1}) (x_{2} - x_{1}) + \dots + (1 - \sum_{k = 1}^{i - 1} p_{k}) (x_{i} - x_{i - 1}) + (1 - \sum_{k = 1}^{i} p_{k}) (x_{i + 1} - x_{i}) + \dots + p_{n} (x_{n} - x_{n - 1}) + 0 \\ = p_{1} x_{1} + p_{2} x_{2} + \dots + p_{n} x_{n} \\ = \sum_{i = 1}^{n} p_{i} x_{i} \\ = E [x] \end{array}

niuers

2020-03-20 00:00

Exercise 5.3.8

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