Probability – How to understand if an exercise is Poison-Bernouli or e-What are the differences

I have three different exercises and I am confused. If it's Poison or Bernuli or if I've seen them and I know them and I'm reading but I'm not so skilled at getting it. I mean I can not detect the exercise of what it is. Can you share my tip to help me get it? I've used Bernouli in cases where he needs poison and opossite.If you can help me do not solve it but explain to me why is Bernouli where you understand, I will be grateful.
first exercise

There is a market race to sell their cornflakes. A company calls
random numbers in order to promote their cornflakes to
customers.We are sure that this company already has 20% of all
What is the possibility that a telephone interlocutor (means that the work
in this company) who calls customers who need more than 3 calls up to
she will succeed someone who is not a customer? My answer: I do not know
which typology to use

second exercise

An Internet service company 4 servers, and both are
"hosting" of the database and the other 2 our software.
possibility that we have damage on a server next year is 1% .We
take for granted that the possibility of appearing harm in all
the server is independently each.If the Internet company are working
there must be at least 1 server that will have the database and a
server that will make the software. What is the posibility of society
Will he work normally and next year?

probability – PDF mapping in a different space

Whether it's the random vectors …

$$ X = begin {bmatrix} X_1, X_2 end {bmatrix} $$
$$ Y = begin {bmatrix} Y_1, Y_2 end {bmatrix} $$

Let there be the probability density functions $ f_ {X} $ and $ f_ {Y} $ (also written as $ f_ {X_1, X_2} $ and $ f_ {Y_1, Y_2} $ respectively).

Whether it's the transformation $ varphi: mathbb R ^ 2 to mathbb R ^ 2 $, and that's true …

$$ Y = varphi (X) $$

Suppose that there is also an inverse $ varphi ^ {- 1} $.

How can we write $ f_ {X_1, X_2} $ in terms of $ f_ {Y_1, Y_2} $, $ varphi $and or $ varphi ^ {- 1} $?

probability – independent and uniformly distributed random variable

Consider three independent sources uniformly distributed (taking values ​​between $ 0 and $ 1 $) Random variables. What is the probability that the middle of the three values ​​(between the smallest and the largest) is between $ a $ and $ b $ or $ 0 ≤ a <b ≤ 1 $?

$ A) 3 (1 – b) a (b – a) $

$ B) 3 ((b – a) – (b ^ {2} – a ^ {2}) / 2) $

$ C) 6 (1 – b) a (b – a) $

$ D) (1 – b) a (b – a) $

$ E) 6 ((b ^ {2} – a ^ {2}) / 2 – (b ^ {3} – a ^ {3}) / 3) $.


Now, formula for uniform distribution $ int_ {a} ^ {b} xdx = left [ frac{x^{2}}{2} right ]_ {a} ^ {b} $

According to the question average value of $ 3 $ the values ​​are inside $ a $ and $ b. $

So, here are 4 possibilities. Let's say that the numbers are x, y, z.

Now,

Case $ 1: $ Only $ y $ can be inside $ a $ and $ b. $

Case $ 2: $ Only $ x, y $ can be inside $ a $ and $ b. $

Case $ 3: $ Only $ y, z $ can be inside $ a $ and $ b. $

Case $ 4: $ $ x, y, z $ everything can be inside $ a $ and $ b. $

So the probability should be $ frac { frac {^ {4} textrm {C} _ {1} times left (b ^ {2} -a ^ {2} right)} {2}} { frac {^ { 8} textrm {C} _ {1} times left (b ^ {2} -a ^ {2} right)} {2}} $

Where am I wrong?

probability – reduce a subspace with unequal probabilities

I was asked an interesting homework question:

Given three pieces in a box, where two have a probability of heads ( $ p (h) = 0.5 $ ) and the third $ p (h) = frac {1} {3} $. I draw a coin and the result is a tail. What is the probability that I did not remove the unfair piece from the box?

I started to solve this problem by trying to find $ p ($unfair$ | t) $, and builds a sample space. Let $ A, B $ represent the right pieces, and $ C $ represent the unfair. The sample space is:

$$ {A_t, A_h, B_t, B_h, C_t, C_h } $$

With probabilities of

$$ { frac {1} {6}, frac {1} {6}, frac {1} {6}, frac {1} {6}, frac {2} {9}, frac {1} {9} } $$

When trying to find $ p ($unfair$ | t) $ we reduce this sample space to:

$$ {A_t, B_t, C_t } $$

I can not understand what the new probabilities are for this sample space. I think $ C_t $ has a higher probability than the other two, but I can not seem to divide that into three probabilities that add to 1.

How could I find the new probabilities of this subspace? I've tried many tricks, such as multiplying the old probabilities by 2, but the sampling space has a total probability> 1.

Any help is appreciated, gratuities included. I would like to be able to do it myself, so do not give me the answer, please.

geometric probability – Determination of sample spaces for Sylvester's four-point problem

I have no knowledge of geometric probability, so I can apologize if the following is wrong or does not make sense:

Hypotheses:

  • no three sampled points are collinear
  • the geometric probability (naive) in Euclidean space is equal to the ratio of lengths or surfaces
  • probabilities do not change under isometric transformations and / or uniform scaling
  • the probability of a set $ P4: = lbrace p_1, p_2, p_3, p_4 rbrace $ four points being in convex configuration in the Euclidean plane is independent of the order in which the elements are drawn by the sampling process; this in turn means that we assume that
  • $ lbrace p_1, p_2, p_3 rbrace $ looks like the corners of $ T_ {max} $, the triangle of the largest area
  • $ lbrace p_1, p_2, p_3 rbrace $ uniformly drawn from the boundary of their circumscribed circle, which has implications for the probability of meeting $ T_ {max} $ with specific values ​​for the pair of smaller central angles.
  • the smallest circle surrounding $ P4 $ is the unit circle centered at the origin
  • the longest side of $ T_ {max} $ is divided in two by the non-negative part of the x-axis

According to the above assumptions, the probability of encountering a convex quadrilateral is equal to the blue zone divided by the red plus blue zone in the images below:

probability of 4 four points in convex configuration if the largest triangle is acute

the sampling area in case of acute toxicity $ T_ {max} $ is equal to the entire disk of the unit

four point probability in convex configuration if the largest triangle is obtuse

the sampling area in case of acute toxicity $ T_ {max} $ is equal to the disk drive with a notch

The notch in the case of obtuse $ T_ {max} $ is due to the assumption that the first three points look like $ T_ {max} $ which implies that the points apart $ T_ {max} $ that generate a deltoid configuration would be in contradiction with the maximality of the zone of $ T_ {max} $

If the above makes sense, the probability that four points are in a convex configuration can be calculated by integrating the blue zone ratios over the entire sampling area defined by the $ T_ {max} $ multiplied by the probability of $ T_ {max} $ resulting in uniform sampling at the boundary of the unit circle.

Questions:

  • Have similar ways of defining sample spaces for Sylvester's four-point problem already been described or studied?
  • what are the objections to the proposed proposal compared to the proposed definition of the available sampling space on $ T_ {max} $?

Note:

in case of obtuse $ T_ {max} $ the surface of the sample space can be calculated based on the angles $ alpha $ and $ beta $ which are adjacent to the longer side of $ T_ {max} $ as follows, keeping in mind that this longest edge is the diameter of the unit circle:

  • the surface of the lower half-blue disk is equal to $ frac { pi} {2} $
  • The area $ A _ { alpha} $ of the union of $ T_ {max} $ with the blue region opposite to the angle $ alpha $ equals $ alpha + sin ( alpha) cos ( alpha) $ and the like $ A _ { beta} $for angle $ beta $
  • the domain of $ A_ {T_ {max}} $ of $ T_ {max} $ equals $ frac {1} { cot ( alpha) + cot ( beta)} $

The area of ​​the sampling area for obtuse $ T_ {max} $ then equal $ frac { pi} {2} + A _ { alpha} + A _ { beta} -A_ {T_ {max}} $

probability or statistics – How to get the correct interpolation for the following data?

I have a dataset DistrTemp.dat. It represents a set of data in the form {{x, y, z, f[x,y,z]}}. Here, x defines a system, while F[x,y,z] is a probability of finding a system in the neighborhood of y, z and has a property
$$
int limits_ {0} ^ { pi} dy int limits {x} ^ {3500} dz f (x, y, z) = 1
$$

My goal is to get a continuous function F[x,y,z] that I can use for integration measurement.

I'm trying to interpolate the dataset for particular values ​​of x:

setDirectory[NotebookDirectory[]]DistrData = Import["DistrTemp.dat","Table"];
DistrInterpolation005[[Theta]HIS_]=
Interpolation[{Log10[#[[2]]], Log10[#[[3]]], #[[4]]} & / @
To select[DistrData, #[[1]]== 0.05 &], InterpolationOrder -> 1][
log10[[Theta]S], Log10[ES]]

Next, I compare the interpolation obtained with the initial dataset:

Plot3D[DistrInterpolation005[DistrInterpolation005[DistrInterpolation005[DistrInterpolation005[[Theta]S, 10 ^ (ES)], {[Theta]S, 0,
Pi}, {ES, Log10[0.05], Log10[3500]}, Image size -> Large,
PlotRange -> All]ListPlot3D[{#[[2]], Log10[#[[3]]], #[[4]]} & / @
To select[DistrData, #[[1]]== 0.05 &], PlotRange -> All,
PlotStyle -> Directive[Red], ImageSize -> Great]

I find that interpolation has been poorly done because some areas are lost, while others are overestimated:

enter the description of the image here

I've tried using WeightedData:

TableCoordinates = Select[DistrData, #[[1]]== 0.05 &][[All, {2, 3}]]Weight of the table = Flatten[DataTemp[[All, 4]]]F1 = WeightedData[TableCoordinates, TableWeights]
DistrWeighted =
Table[f[F1], {f, {HistogramDistribution, SmoothKernelDistribution,
EmpiricalDistribution}}]Table[Plot3D[
PDF[I{[I{[je{[i{[Theta]S, Log10[ES]}], {ES, Log10[0.05],
log10[3500]}, {[Theta]S, 0, Pi},
PlotLabel -> Row[{"Mean: ", Mean[i]}], ImageSize -> Great], {I,
DistrWeighted}]

However, none of the distributions correspond to the desired distribution:

enter the description of the image here

How to properly construct the interpolation?

probability – an infinitely iterative game of giving and withdrawing bills

Imagine a game where, at each stage, a person receives 10 dollar bills (numbered in sequence, starting with 1), then an invoice is removed from the bills that the person has received during the game. Suppose that the game lasts an infinite number of steps. How many bills remains after? As you may have already assumed, the answer depends on how bills are removed.

If the bill with the lowest number is deleted each time, there are no bills left. The deletion of each invoice can be linked to a specific step, so that no invoice can be identified and that it can remain in the collection of the person. There are also other types of deletion that result in a total loss, but this is simple.

If the highest numbered invoice is deleted each time, there will be an infinite number of invoices, because each of the infinite steps gives the person nine bills that will never be deleted. Again, there are other methods to achieve this, but this one is simple.

There are also deletion schemes that one could specify to end up with a finite number of bills, but just wait for the player to receive the correct number of bills before setting up a payout structure. total removal starting at a bill of a specific number. I can give examples later, if necessary.

And if the removal process is randomized? In step 1, invoices 1-10 are awarded and one of these 10 invoices is randomly selected to be withdrawn. In step two, bills 11 to 20 are handed in and one of the 19 remaining bills is randomly selected to be withdrawn. Etc.

What I want is how to determine what is the probability that there are only few or no bills left after an infinite number of steps. I know the infinite monkey theorem, but it is a finite string contained in an infinitely infinite chain and this problem concerns an infinitely infinite chain being part of the set of patterns that remove all the effects, and this difference seems significant. It seems that I can not simply compare the possibilities that have and do not give this result because they are both aleph-1 games, and I have proof that an infinite number of infinite possibilities can not be assumed just balance that way (I do not know if there is one elsewhere, but I've come to this point).

TBH, I have the feeling that I should be able to solve this problem, but I would really appreciate help, because I'm writing my PhD thesis in philosophy and that's a minor point, but I really would like to stay the. possible. Of course, any assistance provided will be quoted (we can discuss how you want this done). If I do it myself (or think I've done it), I'll post a follow up and see what you think.

Thanks in advance!

Probability Theory – $ p_ {y simp} (y neq y) leq p_ {y simp (}} (y ​​ neq y) | | p-p & # 39; | $ for fixed $ p, p & # 39; in[0,1]$

On page 13 of this document: https://arxiv.org/pdf/1708.00489.pdf
it says :

we state claim 1 of Berlind & Urner (2015). For some people $ p, p & # 39; in[0,1]$ and $ y & # 39; in {0,1} $. Then, $$ p_ {y sim p} (y neq y) leq p_ {y sim p} (y neq y)) | | p-p & # 39; |. $$

I looked up and brought it here: https://pdfs.semanticscholar.org/f0f5/e63ca2b8ee32d33c2489225445419364ea99.pdf

But this document does not have the same equation. I want to prove it myself. And I think the notation of the statement is not clear. What is some $ p $, $ p in[0,1]$? I need to help 1) understand the notation of these $ p, p $ and 2) prove the inequality above. Thank you in advance.

probability or statistics – How to optimize the number of bins with weighted data using multiple lists

I try to sort items from multiple lists into buckets containing weighted data. I can do the single list case based on a previous question. Specifically, the code under Update in the first response.

data = {1, 2, 3, 4, 5, 6, 7, 10};
trays = {0, 2, 4, 6, 8, 10};
weight = {0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 1.0};

With[{ranges = Partition[bins, 2, 1]}
Total @ Pick[weights, 
     BitXor[UnitStep[Subtract[data, #1]],
UnitStep[Subtract[data, #2]]], 1]& @@@ ranges]

which gives the following output:
{0,1, 0,5, 0.9, 1.3, 0}

Now I'm trying to generalize at the next entry:

data = {{1, 2, 3, 4, 5, 6, 7, 10}, {3, 5, 6}};
trays = {0, 2, 4, 6, 8, 10};
weight = {{0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 1.0}, {0.4, 0.6, 0.5} };

I can not use the same code because Pick does not process each list sequentially. However, I can get the desired result using this code (where I have the first and second list separately:

With[{ranges = Partition[bins, 2, 1]}
Total @ Pick[weights[[1]],
BITXOR[UnitStep[Subtract[data, #1]],
UnitStep[Subtract[data, #2]]][[1]], 1]& @@@ ranges]With[{ranges = Partition[bins, 2, 1]}
Total @ Pick[weights[[2]],
BITXOR[UnitStep[Subtract[data, #1]],
UnitStep[Subtract[data, #2]]][[2]], 1]& @@@ ranges]

{0,1, 0,5, 0.9, 1.3, 0}
{0, 0.4, 0.6, 0.5, 0}

But I want to generalize to a lot of lists now.
I've tried a loop but when I generalize it loops it does not work:

For[i = 1, 2, i ++,
With[{ranges = Partition[bins, 2, 1]}
Total @ Pick[weights[[i]],
BITXOR[UnitStep[Subtract[data, #1]],
UnitStep[Subtract[data, #2]]][[i]], 1]& @@@ ranges]]

How can I generalize properly?

probability – Derivation of the bootstrap variance of the absolute central moments

The text of my statistics introduces the bootstrap as follows:

  1. Draw $ X_1 ^ {*}, ldots, X_n ^ {*} sim hat {F} _n $.

  2. Calculate $ T_n ^ {*} = g (X_1, ldots, X_n) $.

  3. Repeat steps 1 and 2 $ B $ times to get $ T_ {n, 1} ^ {*}, ldots, T_ {n, B} ^ {*} $.

  4. Let:

$$ v_ {boot} = frac {1} {B} sum_ {b = 1} ^ B (T_ {n, b} ^ {*} –
frac {1} {B} sum_ {r = 1} ^ B T_ {n, r} ^ {*}) ^ 2 $$

Then there is the following exercise:

Let $ T_n = overline {X} _n ^ 2 $, $ mathbb {E} (X_1) = mu $, $ alpha_k =
int left | x- mu right | ^ kdF (x) $
and $ hat { alpha} _k =
n ^ {- 1} sum_ {i = 1} ^ n left | X_i – overline {X} _n right | ^ k $
. CA watch:

$$ v_ {boot} = frac {4 overline {X} _n ^ 2 hat { alpha} _2} {n} +
frac {4 overline {X} _n ^ 2 hat { alpha} _3} {n ^ 2} +
frac { hat { alpha} _4} {n ^ 3} $$

I am confused because I understand $ v_ {boot} $ always be whatever the formula will converge in probability on the variance of the statistic by assuming the empirical CDF with a sufficiently large number of replications (based on WLLN), which means that any formula used must necessarily use replications. But the formula I am asked to derive contains no replication, no variables with an asterisk. Maybe I could interpret that to mean that the replications we draw do not matter, but that seems unlikely. Do I interpret things correctly? Is this an error in the text?