probability: find the p.d.f of $ frac {x_1} { sum_ {i = 1} ^ n {(x_i)}} $ where $ x_i sim Exponetial ( theta) $? and $ x_i $ are i.i.d?

How is the probability density function of the random variable derived? $ frac {x_1} { sum_ {i = 1} ^ n {(x_i)}} $ where $ x_i sim Exponetial ( theta) $? Y $ x_i $ are i.i.d?

I can not think of the bivariate / multivariate transformation needed to do this.

I know how to expand the denominator, but I can not see the solution after expanding the denominator.

reference request – Formula for the volume of $ {x_i in [-N,N]: sum_ {i = 1} ^ n x_i = 0 } $

I'm not an expert in convex geometry but if we define $ a_i sim mathcal {U} ([-N,N]$ where $[-N,N] subset mathbb {R} $ Y $ S_n = sum_ {i = 1} ^ n a_i $ I would like to know if for arbitrary $ N in mathbb {R} _ + $:

  1. The following limit is always true:

begin {equation}
lim_ {n to infty} P (S_n = 0) = 0 tag {*}
end {equation}

  1. There is a simple formula for the volume:

begin {equation}
Vol ( {x_i in [-N,N]: sum_ {i = 1} ^ n x_i = 0 })
end {equation}

So far I have managed to approach the discrete case by modeling it as a random walk in $ mathbb {Z} $. Basically, I was able to prove that if we assume $ a_i sim mathcal {U} ([-N,N]$ where $[-N,N] subset mathbb {Z} $ so:

begin {equation}
lim_ {n to infty} P (S_n = 0) = 0 tag {1}
end {equation}

begin {equation}
lim_ {n a infty} P (| S_n | leq N) = 0 tag {2}
end {equation}

begin {equation}
lim_ {n a infty} P (| S_n |> N) = 1 tag {3}
end {equation}

A description of my analysis is in my blog, Kepler Lounge. Having said that, I very much doubt that he is the first person to make this discovery.

I suspect that the result I am looking for is known to experts in convex analysis, but I do not know what references I should consult.

functional analysis – $ conv (A) = { sum_ {i = 1} ^ n lambda _i x_i: sum_ {i = 1} ^ n lambda_i = 1, lambda_i ge 0, x_i in A, i = 1, cdots, n } $

Leave $ X $ be a normed vector space and $ A $ be a subset of $ X $. $ conv (A) $
It is called the intersection of all convex subsets of $ X $ that contain
$ A $

a) Show that conv (A) is a convex set

b) show that

$$ conv (A) = { sum_ {i = 1} ^ n lambda _i x_i: sum_ {i = 1} ^ n lambda_i = 1,
lambda_i ge 0, x_i in A, i = 1, cdots, n } $$

c) yes $ A $ it is compact then it is $ conv (A) $ compact?

d) Show that if $ A subseteq mathbb {R} ^ n $ it is compact then $ conv (A) $ is
compact

a) Take two elements $ a $ Y $ b $ at the intersection of all convex subsets of $ X $ that contain $ A $. Now take $ lambda a + (1- lambda) b $. It is contained in each of the subsets of $ X $ that contain $ A $, therefore it is contained in the intersection. Q.E.D.

second)

$ leftarrow $:

Suppose by induction that $ sum_ {i = 1} ^ n lambda_i x_i $ It belongs to conv (A). We must prove that $ sum_ {i = 1} ^ {k + 1} lambda_i x_i $, with $ sum_ {i = 1} ^ {k + 1} = 1 $ It also belongs.

$$ sum_ {i = 1} ^ {k + 1} lambda_i x_i = sum_ {i = 1} ^ {k} lambda_i x_i + lambda_ {k + 1} x_ {k + 1} = sum_ {i = 1} ^ {k} lambda_i x_i + (1- sum_ {i = 1} ^ n lambda_i) x_ {k + 1} $$

Now choose $ delta $ such that $ delta sum_ {i = 1} ^ {k} lambda_i = 1 $, so

$$ frac { delta} { delta} left ( sum_ {i = 1} ^ {k} lambda_i x_i + (1- sum_ {i = 1} ^ n lambda_i) x_ {k + 1 } right = = left ( frac {1} { delta} sum_ {i = 1} ^ {k} ( delta lambda_i) x_i + frac {1} { delta} ( delta-1) x_ {k + 1} right) = \ frac {1} { delta} x + left (1- frac {1} { delta} right) x_ {k + 1} $$

which is a collection of elements of $ A $ that adds up to $ 1 $

$ rightarrow $:

I can only say that $ x in conv (A) $, so $ x = 1x + 0 cdot all $ Therefore, it is a combination that adds to $ 1 $ of elements of $ A $?

by do), give me a clue: show that conv (A union B) is the image through a continuous function of the compact $ {( alpha, beta; alpha, beta ge 0, alpha + beta = 1) } times A times B $. I do not understand how to show this.

re) Some clue?

automatons – Is this language free of context? $ Sigma $ = {a, b, #} L = {x1 # x2 # … # xk: k $ geq $ 2, each $ x_i in $ {a, b} * and xi $ neq $ xj for each pair i $ neq $ j}

This question already has an answer here:

Is this language free of context? $ Sigma $ = {a, b, #}, L = {x1 # x2 # … # xk: k$ geq $2, each $ x_i in $ {a, b} * and xi $ neq $ xj for each pair i $ neq $ j} I do not think so, because the PDA can not memorize every word so that it is not equal to every other word.

pr.probability: maximum of sums of iid $ X_i $ & # 39; s where $ X_i $ is the difference of two exponential r.v

Dice $ X_i = A_i – B_i $ where $ A_i sim text {Exp} ( alpha) $ Y $ B_i sim text {Exp} ( lambda) $. Define $ S_k = sum_ {i = 1} ^ k X_i $ with $ S_0 = 0 $Y
$$ M_n = max_ {1 leq k leq n} S_k. $$
Is it possible to calculate the amount? $ mathbb {P} (M_n leq x) $ explicitly? I've tried and the result is written in the bottom, but it's not very explicit …

In Feller's theory of Introduction to probability and its application, he made the observation several times that this type of distribution with two exponential tails is a rare but important case in which almost all the calculations related to the random walk can be made explicit. . (V1.8 Example (b) page 193; XII.2 Example (b) page 395; XII.3 Example (b) page 401) Unfortunately, I could not find any detailed calculations in the book.

A second reference I looked at is paper. "On the distribution of the maximum of sums of random variables mutually independent and identically distributed"by Lajos Takacs (adv. Appl. Prob. 1970) Takacs mentioned that in some particular cases we can calculate $ mathbb {P} (M_n leq x) $ easily. Following his example on page 346 (where he only assumed $ X_i = A_i – B_i $ where $ B_i $ it is exponential and $ A_i $ is not negative), I was able to calculate $ A_i sim text {Exp} ( alpha) $, $ B_i sim text {Exp} ( lambda) $, we have
$$ U (s, p) = sum_ {n = 0} ^ infty mathbb {E} left[e^{-sM_n}right]p ^ n = frac { lambda – frac {s lambda} { gamma (p)}} { lambda – s – frac { lambda p} { alpha + s}} $$
where $ gamma (p) = frac { lambda – alpha + sqrt {( alpha + lambda) ^ 2 – 4 alpha lambda p}} {2} $, a zero of the previous denominator. Is there a way to simplify this to get some kind of explicit formula for $ mathbb {P} (M_n leq x) $?