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Diffstat (limited to 'sem6/prob/m2')
| -rw-r--r-- | sem6/prob/m2/noter.tex | 48 | ||||
| -rw-r--r-- | sem6/prob/m2/opgaver.md | 30 |
2 files changed, 46 insertions, 32 deletions
diff --git a/sem6/prob/m2/noter.tex b/sem6/prob/m2/noter.tex index 3eb2e4f..c35f52b 100644 --- a/sem6/prob/m2/noter.tex +++ b/sem6/prob/m2/noter.tex @@ -1,6 +1,5 @@ \title{Noter til probability m2} -\section{Random Variables} Her mapper man fra et sample space S til en variabel. Her kalder man variablen et stort tal R eller sådan noget. @@ -15,7 +14,7 @@ P(X = x) = 0 $$ -\subsection{Cumulative Distribution Function} +\section{Cumulative Distribution Function} Her måler man prob for at ens random er mindre end et bestemt tal. @@ -33,7 +32,7 @@ Ved discrete random variables vil denne være en slags trappe. Kan sige at den er \emph{continues from the right} eftersom man har $\leq$ i definition. -\subsection{Probability Mass Function} +\section{Probability Mass Function} Works only for discrete random variables. Is defines as the probability that $X = a$: @@ -49,15 +48,17 @@ F(a) = \sum_{all x \leq a} p(a) $$ -\subsection{Probability Density Function} +\section{Probability Density Function} Her finder man P i et evigt lille interval: -Is the derivative of the CDF. +In the following formula PDF is $f$. +\begin{equation*} + \begin{split} + F(a) = P(X \in (-\infty,a]) = \int_{-\infty}^a f(x) dx \\ + f(a) = \frac{d}{da} F(a) + \end{split} +\end{equation*} -$$ -F(a) = P(X \in (-\infty,a]) = \int_{-\infty}^a f(x) dx \\ -f(a) = \frac{d}{da} F(a) -$$ The following must be true: @@ -65,7 +66,7 @@ $$ \int_{-\infty}^{\infty} f(x) dx = 1 $$ -\subsection{Multiple Random Variables} +\section{Multiple Random Variables} Have multiple random variables, which can be or is not correlated. Can define the joined CDF: @@ -81,15 +82,28 @@ F_X(x) = P(X \leq x) = P(X \leq, Y < \infty) = F(x, \infty) $$ One can not go from marginal to the joined, as they do not contain enough information. -This is only possible if X and Y are \emph{independent}. -$$ -F_{XY}(x,y) = F_X(x) \cdot F_Y(x) \\ -p(x,y) = p_X(x) \cdot p_Y(y) \\ -f(x,y) = f_X(x) \cdot f_Y(y) -$$ +However if two random variables, and $A$ and $B$ are two sets of real numbers: +\[ + P(X \in A, Y \in B) = P(X \in A) P(Y \in B)\,. +\] +% This is only possible if X and Y are \emph{independent}. +% \begin{align*} +% F_{XY}(x,y) &= F_X(x) \cdot F_Y(x) \\ +% p(x,y) &= p_X(x) \cdot p_Y(y) \\ +% f(x,y) &= f_X(x) \cdot f_Y(y) +% \end{align*} + +\section{Conditional PDF} + +If $X$ and $Y$ have a joint PDF, then the conditional PDF of X given that $Y=y$ is +\[ + F_{X|Y}(x|y) = \frac {f(x, y)} {f_Y(y)} +\] + +There is also one for PMF not listed here. -\subsection{Joined PMF} +\section{Joined PMF} $$ P_{XY}(x,y) = P(X = x, Y = y) diff --git a/sem6/prob/m2/opgaver.md b/sem6/prob/m2/opgaver.md index 0ce9c77..601aa86 100644 --- a/sem6/prob/m2/opgaver.md +++ b/sem6/prob/m2/opgaver.md @@ -5,16 +5,16 @@ Man kan sige at chances for at en kvinde kommer først er 50%. $$ -P(1) = \frac 1 2 +P(1) = \frac 1 2 = 0.5 $$ -Herefter kræver det at en mand for først og en kvinde får næste. +Herefter kræver det at en mand får først og en kvinde får næste. -$$ -P(2) = \frac 5 10 \cdot \frac 5 9 \\ -P(3) = \frac 5 10 \cdot \frac 4 9 \frac 5 8 \\ -P(4) = \frac 5 10 \cdot \frac 4 9 \frac 3 8 \cdot \frac 5 8 -$$ +\begin{align*} + P(2) &= \frac 5 {10} \cdot \frac 5 9 = 0.2778\\ + P(3) &= \frac 5 {10} \cdot \frac 4 9 \cdot \frac 5 8 = 0.1389 \\ + P(4) &= \frac 5 {10} \cdot \frac 4 9 \cdot \frac 3 8 \cdot \frac 5 7 = 0.0595 +\end{align*} osv. @@ -52,22 +52,22 @@ $$ Først skal man finde $\lambda$. -$$ - \int_{0}^{\infty} \lambda e^{- \frac x {100}} \mathrm{dx} = 1 \\ - \left[ - \lambda 100 \cdot e^{- \frac x {100}}\right]_{0}^{\infty} = 1 \\ - \lambda \cdot 100 = 1 \\ - \lambda = \frac 1 {100} -$$ +\begin{align*} + \int_{0}^{\infty} \lambda e^{- \frac x {100}} \mathrm{dx} &= 1 \\ + \left[ - \lambda 100 \cdot e^{- \frac x {100}}\right]_{0}^{\infty} &= 1 \\ + \lambda \cdot 100 &= 1 \\ + \lambda &= \frac 1 {100} +\end{align*} Nu kan man sætte 50 til 150 ind. $$ - P(50 < x \leq 150) = \int_{50}^{150} f(x) \mathrm{dx} = - e^{- \frac {150} {100}} + e^{ - \frac {50} {100}} = 0.3834 + P(50 < x \leq 150) = \int_{50}^{150} f(x) \mathrm{dx} = \left[ - 100 \cdot \frac 1 {100} \cdot e^{- \frac x {100}} \right]_{50}^{150} = -e^{- \frac{150} {100}} + e^{ - \frac {50} {100}} \approx 0.3834 $$ Derefter kan vi tage fra 0 til 100. $$ - P(x < 100) = \int_{0}^{100} f(x) \mathrm{dx} = - e^{- \frac {100} {100}} = - \frac 1 e + P(x < 100) = \int_{0}^{100} f(x) \mathrm{dx} = - e^{- \frac {100} {100}} + 1 = 1 - \frac 1 e \approx 0.6321 $$ |