ProbabilityDiscussion - Revision history

Admin at 23:53, 3 July 2017

2017-07-03T23:53:42Z

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	[[Category:Math]]		[[Category:Math]]
			[[Category:Probability]]
			[[Category:Statistics]]
			[[Category:Turbulence]]

Admin: /* Aside */

2017-07-03T23:51:43Z

Aside

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	<math>		<math>
	f_{X_1 , X_2 }		f_{X_1 , X_2 }
	<math>		</math>

	and we can get the covariance <math>Cov(X_1 , X_2)</math> and the correlation function <math>\rho_{X_1, X_2}</math>		and we can get the covariance <math>Cov(X_1 , X_2)</math> and the correlation function <math>\rho_{X_1, X_2}</math>

Admin: /* Wiener Process */

2017-07-03T23:49:29Z

Wiener Process

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	<math>		<math>
	\lim_{\Delta t \rightarrow 0} \frac{ X(t_{n+1}) - X(t_n) }{ \Delta t } = \frac{1}{sqrt{ \Delta t} } F(x)		\lim_{\Delta t \rightarrow 0} \frac{ X(t_{n+1}) - X(t_n) }{ \Delta t } = \frac{1}{\sqrt{( \Delta t )} } F(x)
	</math>		</math>

Admin: /* Wiener Process */

2017-07-03T23:48:36Z

Wiener Process

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	</math>		</math>

	where <math>N_{t}^{t+1}</math> is a transition probability - transitioning from <math>t</math> to </math>t+dt</math>		where <math>N_{t}^{t+1}</math> is a transition probability - transitioning from <math>t</math> to <math>t+dt</math>

	This is an initial value problem (think explicit Euler...)		This is an initial value problem (think explicit Euler...)
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	http://files.charlesmartinreid.com/Einstein_BrownianMotion_1905.pdf		http://files.charlesmartinreid.com/Einstein_BrownianMotion_1905.pdf


	==== Ornstein-Uhlenbeck Process ====		==== Ornstein-Uhlenbeck Process ====

Admin: /* Homework 4 */

2017-07-03T23:46:40Z

Homework 4

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	== Homework 4 ==		== Homework 4 ==

	For two ''independent'' random variables X and Y with moment generating functions <math>M_X(t)</math> and M_Y(t), show that the sum of these variables, Z = X+ Y, has a moment generating function <math>M_Z(t) = M_X(t)M_Y(t)</math>. Suggestion: if you first show that for ''independent'' variables <math>E(g(X)h(Y)) = E(g(X))*E(h(Y))</math> then you can use the "cool" way of writing the moment generating function to give this proof in about four steps.		For two ''independent'' random variables X and Y with moment generating functions <math>M_X(t)</math> and M_Y(t), show that the sum of these variables, Z = X+ Y, has a moment generating function <math>M_Z(t) = M_X(t) M_Y(t)</math>.

			Suggestion: if you first show that for ''independent'' variables <math>E(g(X) h(Y)) = E(g(X)) E(h(Y))</math> then you can use the "cool" way of writing the moment generating function to give this proof in about four steps.

	Given: definition of independence		Given: definition of independence
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	<math>		<math>
	E(g(X) h(Y)) = \int_{-\infty}^\infty \int_{-\infty}^\infty f_{X}*f_{Y} g(X) h(Y) = \int_{-\infty}^\infty f_X g(X) dx \int_{-\infty}^\infty f_Y h(Y) dy = E(g(X)) E(h(Y)) </math>		E(g(X) h(Y)) = \int_{-\infty}^\infty \int_{-\infty}^\infty f_{X} f_{Y} g(X) h(Y) = \int_{-\infty}^\infty f_X g(X) dx \int_{-\infty}^\infty f_Y h(Y) dy = E(g(X)) E(h(Y)) </math>

	Now, let's look at two particular functions for g and h:		Now, let's look at two particular functions for g and h:

Admin: /* Homework 4 */

2017-07-03T23:45:56Z

Homework 4

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	Suppose <math> g(X) = \exp{(tX)} </math> and <math> h(Y) = \exp{(tY)} </math>		Suppose <math> g(X) = \exp{(tX)} </math> and <math> h(Y) = \exp{(tY)} </math>

	<math>E( \exp{(tX))} * E( \exp{ (tY) } ) = E( \exp{(tX)} * \exp{(tY)} ) = E( \exp{(t*(X+Y)) } ) = E( \exp{(tZ)} ) = M_Z(t) </math>		<math>E( \exp{(tX))} E( \exp{ (tY) } ) = E( \exp{(tX)} \exp{(tY)} ) = E( \exp{(t (X+Y)) } ) = E( \exp{(tZ)} ) = M_Z(t) </math>

	= October 25, 2010 =		= October 25, 2010 =

Admin: /* Homework 4 */

2017-07-03T23:45:29Z

Homework 4

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	<math>		<math>
	E(g(X)h(Y)) = \int_{-\infty}^\infty \int_{-\infty}^\infty f_{X,Y}(x,y)g(X)*h(Y) dx		E(g(X) h(Y)) = \int_{-\infty}^\infty \int_{-\infty}^\infty f_{X,Y}(x,y) g(X) h(Y) dx
	</math>		</math>

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	<math>		<math>
	E(g(X)h(Y)) = \int_{-\infty}^\infty \int_{-\infty}^\infty f_{X}f_{Y}g(X)h(Y) = \int_{-\infty}^\infty f_Xg(X) dx \int_{-\infty}^\infty f_Yh(Y) dy = E(g(X))*E(h(Y)) </math>		E(g(X) h(Y)) = \int_{-\infty}^\infty \int_{-\infty}^\infty f_{X}*f_{Y} g(X) h(Y) = \int_{-\infty}^\infty f_X g(X) dx \int_{-\infty}^\infty f_Y h(Y) dy = E(g(X)) E(h(Y)) </math>

	~~then if <math>~~ g~~(X) = exp(tX) </math>~~ and ~~<math>~~ h~~(Y) = exp(tY) </math>~~		Now, let's look at two particular functions for g and h:

	<math>E(exp(tX))E(exp(tY)) = E(exp(tX)exp(tY)) = E(exp(t*(X+Y))) = E(exp(tZ)) = M_Z(t) </math>		Suppose <math> g(X) = \exp{(tX)} </math> and <math> h(Y) = \exp{(tY)} </math>

			<math>E( \exp{(tX))} * E( \exp{ (tY) } ) = E( \exp{(tX)} * \exp{(tY)} ) = E( \exp{(t*(X+Y)) } ) = E( \exp{(tZ)} ) = M_Z(t) </math>

	= October 25, 2010 =		= October 25, 2010 =

Admin: /* Homework Problem 1 */

2017-07-03T23:39:16Z

Homework Problem 1

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	We know <math>x < 2</math>, so it can be 0 or 1, and plugging this into expression for Poisson distribution:		We know <math>x < 2</math>, so it can be 0 or 1, and plugging this into expression for Poisson distribution:

	<math>\frac{ \exp{-\lambda} \lambda^{0} }{ 0! } + \frac{ \exp{-\lambda} \lambda^{1} }{ 1! } < 0.01</math>		<math>\frac{ \exp{(-\lambda)} \lambda^{0} }{ 0! } + \frac{ \exp{(-\lambda)} \lambda^{1} }{ 1! } < 0.01</math>

	(replace the <math><</math> sign with an <math>=</math> sign... and solve for <math>\lambda</math>...)		(replace the <math><</math> sign with an <math>=</math> sign... and solve for <math>\lambda</math>...)

Admin: /* Homework Problem 1 */

2017-07-03T23:38:51Z

Homework Problem 1

← Older revision		Revision as of 23:38, 3 July 2017
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	<math>		<math>
	P(X=x) = \frac{ \~~exp~~{ -\lambda ~~} \lambda^{x~~} }{ x! }		P(X=x) = \frac{ \lambda^{x} e^{-\lambda} }{ x! }
	</math>		</math>

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	\lambda = 6.638		\lambda = 6.638
	</math>		</math>



	== Homework 2 ==		== Homework 2 ==

Admin: /* Part b */

2017-07-03T23:35:09Z

Part b

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	=== Part b ===		=== Part b ===

	Find probability that a random variable x is less than t, e.g. <math>P(x<t) \forall t \in \mathfrak{R}</math>		Find probability that a random variable x is less than t, e.g. <math>P(x<t) \quad \forall t \in \mathfrak{R}</math>

	<math>		<math>