User:IssaRice/Additivity of small risks - Revision history

IssaRice at 20:39, 14 November 2020

2020-11-14T20:39:03Z

← Older revision		Revision as of 20:39, 14 November 2020
Line 4:		Line 4:

	But now let <math>m_i := -\log(1 - p_i)</math>. Then we have <math>\Pr(\mathrm{death}) = 1 - \exp(-\sum_i m_i)</math>. When each <math>p_i</math> is small, we have <math>m_i \approx p_i</math>. Thus <math>\sum_i m_i</math> is also small. For small positive values of <math>x</math> we have <math>1 - \exp(-x) \approx x</math>. Thus we have <math>\Pr(\mathrm{death}) = 1 - \exp(-\sum_i m_i) \approx \sum_i m_i \approx \sum_i p_i</math>.		But now let <math>m_i := -\log(1 - p_i)</math>. Then we have <math>\Pr(\mathrm{death}) = 1 - \exp(-\sum_i m_i)</math>. When each <math>p_i</math> is small, we have <math>m_i \approx p_i</math>. Thus <math>\sum_i m_i</math> is also small. For small positive values of <math>x</math> we have <math>1 - \exp(-x) \approx x</math>. Thus we have <math>\Pr(\mathrm{death}) = 1 - \exp(-\sum_i m_i) \approx \sum_i m_i \approx \sum_i p_i</math>.

			If <math>y = -\log(1-x)</math> then <math>x = 1-\exp(-y)</math> which means that the two functions are inverses of each other. So it makes sense that when either quantity is small, the other is also small.

IssaRice at 22:43, 13 November 2020

2020-11-13T22:43:25Z

← Older revision		Revision as of 22:43, 13 November 2020
Line 1:		Line 1:
	Suppose you want to engage in some risky activities such as driving on the highway or swimming in the river, which have some small probability of resulting in death. It turns out that when the probabilities involved are independent and small, one can simply add the probabilities together instead of doing the more complicated correct calculation.		Suppose you want to engage in some risky activities such as driving on the highway or swimming in the river, which have some small probability of resulting in death. It turns out that when the probabilities involved are independent and small, one can simply add the probabilities together instead of doing the more complicated correct calculation.

	Let <math>p_1, \dots, p_n</math> be the probabilities of the risk of dying from each of activities <math>1, \dots, n</math>, where each <math>0\leq p_i \ll 1</math>. Then the probability of dying from doing all activities is <math>\Pr(\mathrm{death}) = 1 - \prod_i (1-p_i)</math>. This happens because for independent probabilities we know how to take conjunctions (AND) but not disjunctions (OR) so we must first invert and use de Morgan's laws then invert again.		Let <math>p_1, \dots, p_n</math> be the probabilities of the risk of dying from each of activities <math>1, \dots, n</math>, where each <math>0\leq p_i \ll 1</math>. Then the probability of dying from doing all activities is <math>\Pr(\mathrm{death}) = 1 - \prod_i (1-p_i)</math>. This happens because for independent probabilities we know how to take conjunctions (AND) by multiplying but not disjunctions (OR) so we must first invert and use de Morgan's laws then invert again.

	But now let <math>m_i := -\log(1 - p_i)</math>. Then we have <math>\Pr(\mathrm{death}) = 1 - \exp(-\sum_i m_i)</math>. When each <math>p_i</math> is small, we have <math>m_i \approx p_i</math>. Thus <math>\sum_i m_i</math> is also small. For small positive values of <math>x</math> we have <math>1 - \exp(-x) \approx x</math>. Thus we have <math>\Pr(\mathrm{death}) = 1 - \exp(-\sum_i m_i) \approx \sum_i m_i \approx \sum_i p_i</math>.		But now let <math>m_i := -\log(1 - p_i)</math>. Then we have <math>\Pr(\mathrm{death}) = 1 - \exp(-\sum_i m_i)</math>. When each <math>p_i</math> is small, we have <math>m_i \approx p_i</math>. Thus <math>\sum_i m_i</math> is also small. For small positive values of <math>x</math> we have <math>1 - \exp(-x) \approx x</math>. Thus we have <math>\Pr(\mathrm{death}) = 1 - \exp(-\sum_i m_i) \approx \sum_i m_i \approx \sum_i p_i</math>.

IssaRice at 22:42, 13 November 2020

2020-11-13T22:42:37Z

← Older revision		Revision as of 22:42, 13 November 2020
Line 3:		Line 3:
	Let <math>p_1, \dots, p_n</math> be the probabilities of the risk of dying from each of activities <math>1, \dots, n</math>, where each <math>0\leq p_i \ll 1</math>. Then the probability of dying from doing all activities is <math>\Pr(\mathrm{death}) = 1 - \prod_i (1-p_i)</math>. This happens because for independent probabilities we know how to take conjunctions (AND) but not disjunctions (OR) so we must first invert and use de Morgan's laws then invert again.		Let <math>p_1, \dots, p_n</math> be the probabilities of the risk of dying from each of activities <math>1, \dots, n</math>, where each <math>0\leq p_i \ll 1</math>. Then the probability of dying from doing all activities is <math>\Pr(\mathrm{death}) = 1 - \prod_i (1-p_i)</math>. This happens because for independent probabilities we know how to take conjunctions (AND) but not disjunctions (OR) so we must first invert and use de Morgan's laws then invert again.

	But now let <math>m_i := -\log(1 - p_i)</math>. Then we have <math>\Pr(\mathrm{death}) = 1 - \exp(-\sum_i m_i)</math>.		But now let <math>m_i := -\log(1 - p_i)</math>. Then we have <math>\Pr(\mathrm{death}) = 1 - \exp(-\sum_i m_i)</math>. When each <math>p_i</math> is small, we have <math>m_i \approx p_i</math>. Thus <math>\sum_i m_i</math> is also small. For small positive values of <math>x</math> we have <math>1 - \exp(-x) \approx x</math>. Thus we have <math>\Pr(\mathrm{death}) = 1 - \exp(-\sum_i m_i) \approx \sum_i m_i \approx \sum_i p_i</math>.

IssaRice at 22:39, 13 November 2020

2020-11-13T22:39:07Z

← Older revision		Revision as of 22:39, 13 November 2020
Line 1:		Line 1:
	Suppose you want to engage in some risky activities such as driving on the highway or swimming in the river, which have some small probability of resulting in death. It turns out that when the probabilities involved are independent and small, one can simply add the probabilities together instead of doing the more complicated correct calculation.		Suppose you want to engage in some risky activities such as driving on the highway or swimming in the river, which have some small probability of resulting in death. It turns out that when the probabilities involved are independent and small, one can simply add the probabilities together instead of doing the more complicated correct calculation.

	Let <math>p_1, \dots, p_n</math> be the probabilities of the risk of dying from each of activities <math>1, \dots, n</math>, where each <math>0\leq p_i \ll 1</math>. Then the probability of dying from doing all activities is <math>1 - \prod_i (1-p_i)</math>.		Let <math>p_1, \dots, p_n</math> be the probabilities of the risk of dying from each of activities <math>1, \dots, n</math>, where each <math>0\leq p_i \ll 1</math>. Then the probability of dying from doing all activities is <math>\Pr(\mathrm{death}) = 1 - \prod_i (1-p_i)</math>. This happens because for independent probabilities we know how to take conjunctions (AND) but not disjunctions (OR) so we must first invert and use de Morgan's laws then invert again.

			But now let <math>m_i := -\log(1 - p_i)</math>. Then we have <math>\Pr(\mathrm{death}) = 1 - \exp(-\sum_i m_i)</math>.

IssaRice: Created page with "Suppose you want to engage in some risky activities such as driving on the highway or swimming in the river, which have some small probability of resulting in death. It turns..."

2020-11-13T22:35:05Z

Created page with "Suppose you want to engage in some risky activities such as driving on the highway or swimming in the river, which have some small probability of resulting in death. It turns..."

New page

Suppose you want to engage in some risky activities such as driving on the highway or swimming in the river, which have some small probability of resulting in death. It turns out that when the probabilities involved are independent and small, one can simply add the probabilities together instead of doing the more complicated correct calculation.

Let <math>p_1, \dots, p_n</math> be the probabilities of the risk of dying from each of activities <math>1, \dots, n</math>, where each <math>0\leq p_i \ll 1</math>. Then the probability of dying from doing all activities is <math>1 - \prod_i (1-p_i)</math>.