From f03b2f31add52cb4726b7b64d09e1e0466df39e7 Mon Sep 17 00:00:00 2001 From: Julian T Date: Tue, 9 Feb 2021 11:53:14 +0100 Subject: Add notes and assignments for prob m2 --- sem6/prob/m2/noter.md | 102 ++++++++++++++++++++++++++++++++++++++++++++++++ sem6/prob/m2/opgaver.md | 53 +++++++++++++++++++++++++ 2 files changed, 155 insertions(+) create mode 100644 sem6/prob/m2/noter.md create mode 100644 sem6/prob/m2/opgaver.md (limited to 'sem6') diff --git a/sem6/prob/m2/noter.md b/sem6/prob/m2/noter.md new file mode 100644 index 0000000..f8faec4 --- /dev/null +++ b/sem6/prob/m2/noter.md @@ -0,0 +1,102 @@ +# Noter til probability m2 + +## Random variables + +Her mapper man fra et sample space S til en variabel. +Her kalder man variablen et stort tal R eller sådan noget. + +Derfor er et random variabel egentlig en transformation mellem S og real tal. + +*X er en descrete random variable hvis dens range er countable.* + +For continues random variables the following is true: + +$$ +P(X = x) = 0 +$$ + +## Functions beskriver ens random variable + +### Cumulative Distribution function + +Her måler man prob for at ens random er mindre end et bestemt tal. + +$$ +F(x) = P(X \leq x) +$$ + +Man kan også finde det for en range: + +$$ +P(a < X \leq b) = F(b) - F(a) +$$ + +Ved discrete random variables vil denne være en slags trappe. + +Kan sige at den er *continues from the right* eftersom man har $\leq$ i definition. + +### Probability Mass Function + +Works only for discrete random variables. +Is defines as the probability that $X = a$: + +$$ +p(a) = P(X = a) +$$ + +From here CDF can be found: + +$$ + F(a) = \sum_{all x \leq a} p(a) +$$ + + + +### Probability Density Function + +Her finder man P i et evigt lille interval: +Is the derivative of the CDF. + +$$ + F(a) = P(X \in (-\infty,a]) = \int_{-\infty}^a f(x) dx \\ + f(a) = \frac{d}{da} F(a) +$$ + +The following must be true: + +$$ + \int_{-\infty}^{\infty} f(x) dx = 1 +$$ + +## Multiple random variables + +Have multiple random variables, which can be or is not correlated. +Can define the joined CDF: + +$$ + F_{XY}(x,y) = P(X \leq x, Y \leq y) +$$ + +One can also find the probability of one of the variables. (The *marginal*) + +$$ + F_X(x) = P(X \leq x) = P(X \leq, Y < \intfy) = F(x, \infty) +$$ + +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 **independent**. + +$$ + F_XY(x,y) = F_X(x) \cdot F_Y(x) \\ + p(x,y) = p_X(x) \cdot p_Y(y) \\ + f(x,y) = f_X(x) \cdot f_Y(y) +$$ + +### Joined PMF + +$$ + P_{XY}(x,y) = P(X = x, Y = y) +$$ + + vim: spell spelllang=da,en + diff --git a/sem6/prob/m2/opgaver.md b/sem6/prob/m2/opgaver.md new file mode 100644 index 0000000..a5d0cdb --- /dev/null +++ b/sem6/prob/m2/opgaver.md @@ -0,0 +1,53 @@ +# Opgaver til prob m2 + +## Opgave 1 + +Man kan sige at chances for at en kvinde kommer først er 50%. + +$$ +P(1) = \frac 1 2 +$$ + +Herefter kræver det at en mand for først og en kvinde får næste. + +$$ +P(2) = \frac 5 10 \cdot \frac 5 9 \\ +P(3) = \frac 5 10 \cdot \frac 4 9 \frac 5 8 \\ +P(4) = \frac 5 10 \cdot \frac 4 9 \frac 3 8 \cdot \frac 5 8 +$$ + +osv. + +## Opgave 2 + +### A) + +Dette har jeg gjort på papir. + +### B) + +$$ +P\left(X > \frac 1 2\right) = 1 - F\left(\frac 1 2\right) = 1 - \frac 1 4 = \frac 3 4 +$$ + +### C) + +$$ +P(2 < X \leq 4) = F(4) - F(2) = 1 - \frac {11} {12} = \frac 1 {12} +$$ + +### D) + +$$ +P(X < 3) = \frac {11}{12} +$$ + +### E) + +$$ +P(X = 1) = \frac 2 3 - \frac 1 2 = \frac 1 6 +$$ + +## Opgave 3 + + -- cgit v1.2.3