Limiting a sequence of moment generating functions












3














I was trying to solve the following problem:




Let ${X_n}_{n=1}^{infty}$ be a sequence of independent random variables with the probability mass function $P{X_n = pm1 } = frac{1}{2}$, $n in mathbb{N}$. Let $Z_n=sum_{j=1}^{n}{X_j/2^j}$. Show that $Z_n xrightarrow{L} Z$, where $Z sim U[-1, 1]$.




(From An Introduction to Probability and Statistics, V.K. Rohatgi & A. K. Md. Saleh, (c) 2015, Problems 7.5, Page 320)



Here $xrightarrow{L}$ means convergence in law (or in distribution), and $U[-1, 1]$ is the uniform distribution on the interval $[-1, 1]$.



My approach was the following:



We need to show that $$lim_{nrightarrowinfty} M_{Z_n}(t) = M_{Z}(t) = frac{e^{1 times t} - e^{-1 times t}}{t times (1 - (-1))} = frac{e^t - e^{-t}}{2t}.$$
Since
$$M_{Z_n}(t) = E_{Z_n}left(e^{tZ_n}right) = Eleft(e^{tsum_{j = 1}^{n}{frac{X_j}{2^j}}}right) = Eleft(prod_{j=1}^{n}{e^{tfrac{X_j}{2^j}}} right) = prod_{j=1}^{n}{E_{X_j}left(e^{tfrac{X_j}{2^j}} right)},\
E_{X_j}left( e^{t frac{X_j}{2^j}} right) = e^{t times frac{-1}{2^j}} times frac{1}{2} + e^{t times frac{1}{2^j}} times frac{1}{2} = frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right),$$

then
$$M_{Z_n}(t) = prod_{j = 1}^{n}{frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right)}.$$



I cannot see how this sequence of functions converges to the required moment generating function of $U[-1,1]$.



I had many attempts, for instance using the power series representation of $e^x$ and limiting approximations, but failed in them all. After that I started thinking that perhaps I am missing knowledge of some theorems.



Any idea how to proceed?










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  • Why not using characteristic functions?
    – Saad
    3 hours ago










  • @Saad I tried that, but how does it make any difference? I thought that we use the characteristic function only when the moment generating function doesn't exist. However, here the moment generating function exists and is nice.
    – Noor AlYaqeen
    3 hours ago


















3














I was trying to solve the following problem:




Let ${X_n}_{n=1}^{infty}$ be a sequence of independent random variables with the probability mass function $P{X_n = pm1 } = frac{1}{2}$, $n in mathbb{N}$. Let $Z_n=sum_{j=1}^{n}{X_j/2^j}$. Show that $Z_n xrightarrow{L} Z$, where $Z sim U[-1, 1]$.




(From An Introduction to Probability and Statistics, V.K. Rohatgi & A. K. Md. Saleh, (c) 2015, Problems 7.5, Page 320)



Here $xrightarrow{L}$ means convergence in law (or in distribution), and $U[-1, 1]$ is the uniform distribution on the interval $[-1, 1]$.



My approach was the following:



We need to show that $$lim_{nrightarrowinfty} M_{Z_n}(t) = M_{Z}(t) = frac{e^{1 times t} - e^{-1 times t}}{t times (1 - (-1))} = frac{e^t - e^{-t}}{2t}.$$
Since
$$M_{Z_n}(t) = E_{Z_n}left(e^{tZ_n}right) = Eleft(e^{tsum_{j = 1}^{n}{frac{X_j}{2^j}}}right) = Eleft(prod_{j=1}^{n}{e^{tfrac{X_j}{2^j}}} right) = prod_{j=1}^{n}{E_{X_j}left(e^{tfrac{X_j}{2^j}} right)},\
E_{X_j}left( e^{t frac{X_j}{2^j}} right) = e^{t times frac{-1}{2^j}} times frac{1}{2} + e^{t times frac{1}{2^j}} times frac{1}{2} = frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right),$$

then
$$M_{Z_n}(t) = prod_{j = 1}^{n}{frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right)}.$$



I cannot see how this sequence of functions converges to the required moment generating function of $U[-1,1]$.



I had many attempts, for instance using the power series representation of $e^x$ and limiting approximations, but failed in them all. After that I started thinking that perhaps I am missing knowledge of some theorems.



Any idea how to proceed?










share|cite|improve this question
























  • Why not using characteristic functions?
    – Saad
    3 hours ago










  • @Saad I tried that, but how does it make any difference? I thought that we use the characteristic function only when the moment generating function doesn't exist. However, here the moment generating function exists and is nice.
    – Noor AlYaqeen
    3 hours ago
















3












3








3


1





I was trying to solve the following problem:




Let ${X_n}_{n=1}^{infty}$ be a sequence of independent random variables with the probability mass function $P{X_n = pm1 } = frac{1}{2}$, $n in mathbb{N}$. Let $Z_n=sum_{j=1}^{n}{X_j/2^j}$. Show that $Z_n xrightarrow{L} Z$, where $Z sim U[-1, 1]$.




(From An Introduction to Probability and Statistics, V.K. Rohatgi & A. K. Md. Saleh, (c) 2015, Problems 7.5, Page 320)



Here $xrightarrow{L}$ means convergence in law (or in distribution), and $U[-1, 1]$ is the uniform distribution on the interval $[-1, 1]$.



My approach was the following:



We need to show that $$lim_{nrightarrowinfty} M_{Z_n}(t) = M_{Z}(t) = frac{e^{1 times t} - e^{-1 times t}}{t times (1 - (-1))} = frac{e^t - e^{-t}}{2t}.$$
Since
$$M_{Z_n}(t) = E_{Z_n}left(e^{tZ_n}right) = Eleft(e^{tsum_{j = 1}^{n}{frac{X_j}{2^j}}}right) = Eleft(prod_{j=1}^{n}{e^{tfrac{X_j}{2^j}}} right) = prod_{j=1}^{n}{E_{X_j}left(e^{tfrac{X_j}{2^j}} right)},\
E_{X_j}left( e^{t frac{X_j}{2^j}} right) = e^{t times frac{-1}{2^j}} times frac{1}{2} + e^{t times frac{1}{2^j}} times frac{1}{2} = frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right),$$

then
$$M_{Z_n}(t) = prod_{j = 1}^{n}{frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right)}.$$



I cannot see how this sequence of functions converges to the required moment generating function of $U[-1,1]$.



I had many attempts, for instance using the power series representation of $e^x$ and limiting approximations, but failed in them all. After that I started thinking that perhaps I am missing knowledge of some theorems.



Any idea how to proceed?










share|cite|improve this question















I was trying to solve the following problem:




Let ${X_n}_{n=1}^{infty}$ be a sequence of independent random variables with the probability mass function $P{X_n = pm1 } = frac{1}{2}$, $n in mathbb{N}$. Let $Z_n=sum_{j=1}^{n}{X_j/2^j}$. Show that $Z_n xrightarrow{L} Z$, where $Z sim U[-1, 1]$.




(From An Introduction to Probability and Statistics, V.K. Rohatgi & A. K. Md. Saleh, (c) 2015, Problems 7.5, Page 320)



Here $xrightarrow{L}$ means convergence in law (or in distribution), and $U[-1, 1]$ is the uniform distribution on the interval $[-1, 1]$.



My approach was the following:



We need to show that $$lim_{nrightarrowinfty} M_{Z_n}(t) = M_{Z}(t) = frac{e^{1 times t} - e^{-1 times t}}{t times (1 - (-1))} = frac{e^t - e^{-t}}{2t}.$$
Since
$$M_{Z_n}(t) = E_{Z_n}left(e^{tZ_n}right) = Eleft(e^{tsum_{j = 1}^{n}{frac{X_j}{2^j}}}right) = Eleft(prod_{j=1}^{n}{e^{tfrac{X_j}{2^j}}} right) = prod_{j=1}^{n}{E_{X_j}left(e^{tfrac{X_j}{2^j}} right)},\
E_{X_j}left( e^{t frac{X_j}{2^j}} right) = e^{t times frac{-1}{2^j}} times frac{1}{2} + e^{t times frac{1}{2^j}} times frac{1}{2} = frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right),$$

then
$$M_{Z_n}(t) = prod_{j = 1}^{n}{frac{1}{2} left( e^{frac{t}{2^j}} + e^{frac{-t}{2^j}} right)}.$$



I cannot see how this sequence of functions converges to the required moment generating function of $U[-1,1]$.



I had many attempts, for instance using the power series representation of $e^x$ and limiting approximations, but failed in them all. After that I started thinking that perhaps I am missing knowledge of some theorems.



Any idea how to proceed?







probability-theory probability-limit-theorems






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edited 2 hours ago









Saad

19.7k92252




19.7k92252










asked 3 hours ago









Noor AlYaqeen

215




215












  • Why not using characteristic functions?
    – Saad
    3 hours ago










  • @Saad I tried that, but how does it make any difference? I thought that we use the characteristic function only when the moment generating function doesn't exist. However, here the moment generating function exists and is nice.
    – Noor AlYaqeen
    3 hours ago




















  • Why not using characteristic functions?
    – Saad
    3 hours ago










  • @Saad I tried that, but how does it make any difference? I thought that we use the characteristic function only when the moment generating function doesn't exist. However, here the moment generating function exists and is nice.
    – Noor AlYaqeen
    3 hours ago


















Why not using characteristic functions?
– Saad
3 hours ago




Why not using characteristic functions?
– Saad
3 hours ago












@Saad I tried that, but how does it make any difference? I thought that we use the characteristic function only when the moment generating function doesn't exist. However, here the moment generating function exists and is nice.
– Noor AlYaqeen
3 hours ago






@Saad I tried that, but how does it make any difference? I thought that we use the characteristic function only when the moment generating function doesn't exist. However, here the moment generating function exists and is nice.
– Noor AlYaqeen
3 hours ago












3 Answers
3






active

oldest

votes


















1














$defi{mathrm{i}}defd{mathrm{d}}$Since $X_1, X_2, cdots$ are independent and $displaystyle Z_n = sumlimits_{k = 1}^n frac{X_k}{2^k}$, then$$
φ_{Z_n}(t) = prod_{k = 1}^n φ_{X_k}left( frac{t}{2^k} right) = prod_{k = 1}^n cosleft( frac{t}{2^k} right) = frac{sin t}{2^n sinleft( dfrac{t}{2^n} right)}, quad forall t in mathbb{R}^*
$$

which implies $displaystyle lim_{n → ∞} φ_{Z_n}(t) = frac{sin t}{t}$, and the limit is the characteristic function of $Z sim U(-1, 1)$. By the continuity theorem, $Z_n xrightarrow{mathrm{d}} Z$.






share|cite|improve this answer





















  • I didn't know you could give more than one answer...
    – maridia
    1 hour ago










  • @Saad what was the identity you used to derive $prod_{k=1}^{n}{cosleft( frac{t}{2^k} right)} = frac{sin t}{2^n sinleft(frac{t}{2^n} right)}$?
    – Noor AlYaqeen
    33 mins ago





















1














$defe{mathrm{e}}$Another method: Note that $a + b = dfrac{a^2 - b^2}{a - b}$ for $a ≠ b$, then$$
M_{Z_n}(t) = frac{1}{2^n} prod_{k = 1}^n left( expleft( frac{t}{2^k} right) + expleft( -frac{t}{2^k} right) right) = frac{1}{2^n} · frac{e^t - e^{-t}}{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}.
$$

Because$$
lim_{n → ∞} frac{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}{dfrac{t}{2^n} - left( -dfrac{t}{2^n} right)} = (e^x)'bigr|_{x = 0} = 1,
$$

then $limlimits_{n → ∞} M_{Z_n}(t) = dfrac{1}{2t} (e^t - e^{-t})$ and the result follows.






share|cite|improve this answer





















  • Thanks this is the best answer!
    – Noor AlYaqeen
    9 mins ago



















1














In addition to characteristic functions, one may also approach the problem via binary representation of integers, which is not as short as the answers with characteristic functions, but is quite straightforward.



Indeed, rewrite
$$
Z_n = frac{1}{2^n} sumlimits_{j=1}^n 2^{n-j} X_j : = frac{1}{2^n}S_n.
$$

Define $Lambda_{+} = { 1leq j leq 2^n: X_j = 1 }$, and let $Lambda_-$ be the complement of $Lambda_+$ in $1leq jleq 2^n$. Then,
$$
S_n = sum_{jin Lambda_+} - sum_{j in Lambda_-} = sum_{jin Lambda_+} - left( 2^n - 1 - sumlimits_{jin Lambda_+} right) = 2sumlimits_{j in Lambda_+} 2^{n-j} - (2^n - 1) tag{1}.
$$

Thus, with $S_n$ we cover all integers from $-(2^n - 1) , ... 2^n - 1$ of the form $(1)$, which are precisely all the odd integers from $-(2^n - 1), ... (2^n - 1)$, $2^n$ in total. Thus, if $i in [-(2^n - 1), ..., 2^n - 1] $ is even then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 0 tag{2}
$$

and if $i $ is odd, then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 2^{-n}, tag{3}
$$

since there is a single choice of index set $Lambda_+$ in $(1)$, and hence $(3)$ follows in view of independence of ${X_j}$.



From $(2)$ and $(3)$ we see, by counting the number of odd integers, that for any integer $-2^{n-1} + 1 leq i leq 2^{n-1}$ one has
$$
mathbb{P}left(Z_n leq frac{2i - 1}{2^n} right) = frac{1}{2} + frac{i}{2^n}.
$$



It follows that the distribution function $F_n$ of $Z_n$ coincides, on odd dyadic rationals from $[-1,1]$ (and obviously everywhere on $(-infty, -1] cup [1,infty)$) with the distribution function $F$ of a random variable with the law $U[-1, 1]$. The density of dyadic rationals and right-continuity of cdf imply $F_n to F$ everywhere on $[-1,1]$, hence the claim.






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  • Thank you, a very interesting solution! although, a bit complicated.
    – Noor AlYaqeen
    21 mins ago











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3 Answers
3






active

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3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














$defi{mathrm{i}}defd{mathrm{d}}$Since $X_1, X_2, cdots$ are independent and $displaystyle Z_n = sumlimits_{k = 1}^n frac{X_k}{2^k}$, then$$
φ_{Z_n}(t) = prod_{k = 1}^n φ_{X_k}left( frac{t}{2^k} right) = prod_{k = 1}^n cosleft( frac{t}{2^k} right) = frac{sin t}{2^n sinleft( dfrac{t}{2^n} right)}, quad forall t in mathbb{R}^*
$$

which implies $displaystyle lim_{n → ∞} φ_{Z_n}(t) = frac{sin t}{t}$, and the limit is the characteristic function of $Z sim U(-1, 1)$. By the continuity theorem, $Z_n xrightarrow{mathrm{d}} Z$.






share|cite|improve this answer





















  • I didn't know you could give more than one answer...
    – maridia
    1 hour ago










  • @Saad what was the identity you used to derive $prod_{k=1}^{n}{cosleft( frac{t}{2^k} right)} = frac{sin t}{2^n sinleft(frac{t}{2^n} right)}$?
    – Noor AlYaqeen
    33 mins ago


















1














$defi{mathrm{i}}defd{mathrm{d}}$Since $X_1, X_2, cdots$ are independent and $displaystyle Z_n = sumlimits_{k = 1}^n frac{X_k}{2^k}$, then$$
φ_{Z_n}(t) = prod_{k = 1}^n φ_{X_k}left( frac{t}{2^k} right) = prod_{k = 1}^n cosleft( frac{t}{2^k} right) = frac{sin t}{2^n sinleft( dfrac{t}{2^n} right)}, quad forall t in mathbb{R}^*
$$

which implies $displaystyle lim_{n → ∞} φ_{Z_n}(t) = frac{sin t}{t}$, and the limit is the characteristic function of $Z sim U(-1, 1)$. By the continuity theorem, $Z_n xrightarrow{mathrm{d}} Z$.






share|cite|improve this answer





















  • I didn't know you could give more than one answer...
    – maridia
    1 hour ago










  • @Saad what was the identity you used to derive $prod_{k=1}^{n}{cosleft( frac{t}{2^k} right)} = frac{sin t}{2^n sinleft(frac{t}{2^n} right)}$?
    – Noor AlYaqeen
    33 mins ago
















1












1








1






$defi{mathrm{i}}defd{mathrm{d}}$Since $X_1, X_2, cdots$ are independent and $displaystyle Z_n = sumlimits_{k = 1}^n frac{X_k}{2^k}$, then$$
φ_{Z_n}(t) = prod_{k = 1}^n φ_{X_k}left( frac{t}{2^k} right) = prod_{k = 1}^n cosleft( frac{t}{2^k} right) = frac{sin t}{2^n sinleft( dfrac{t}{2^n} right)}, quad forall t in mathbb{R}^*
$$

which implies $displaystyle lim_{n → ∞} φ_{Z_n}(t) = frac{sin t}{t}$, and the limit is the characteristic function of $Z sim U(-1, 1)$. By the continuity theorem, $Z_n xrightarrow{mathrm{d}} Z$.






share|cite|improve this answer












$defi{mathrm{i}}defd{mathrm{d}}$Since $X_1, X_2, cdots$ are independent and $displaystyle Z_n = sumlimits_{k = 1}^n frac{X_k}{2^k}$, then$$
φ_{Z_n}(t) = prod_{k = 1}^n φ_{X_k}left( frac{t}{2^k} right) = prod_{k = 1}^n cosleft( frac{t}{2^k} right) = frac{sin t}{2^n sinleft( dfrac{t}{2^n} right)}, quad forall t in mathbb{R}^*
$$

which implies $displaystyle lim_{n → ∞} φ_{Z_n}(t) = frac{sin t}{t}$, and the limit is the characteristic function of $Z sim U(-1, 1)$. By the continuity theorem, $Z_n xrightarrow{mathrm{d}} Z$.







share|cite|improve this answer












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answered 2 hours ago









Saad

19.7k92252




19.7k92252












  • I didn't know you could give more than one answer...
    – maridia
    1 hour ago










  • @Saad what was the identity you used to derive $prod_{k=1}^{n}{cosleft( frac{t}{2^k} right)} = frac{sin t}{2^n sinleft(frac{t}{2^n} right)}$?
    – Noor AlYaqeen
    33 mins ago




















  • I didn't know you could give more than one answer...
    – maridia
    1 hour ago










  • @Saad what was the identity you used to derive $prod_{k=1}^{n}{cosleft( frac{t}{2^k} right)} = frac{sin t}{2^n sinleft(frac{t}{2^n} right)}$?
    – Noor AlYaqeen
    33 mins ago


















I didn't know you could give more than one answer...
– maridia
1 hour ago




I didn't know you could give more than one answer...
– maridia
1 hour ago












@Saad what was the identity you used to derive $prod_{k=1}^{n}{cosleft( frac{t}{2^k} right)} = frac{sin t}{2^n sinleft(frac{t}{2^n} right)}$?
– Noor AlYaqeen
33 mins ago






@Saad what was the identity you used to derive $prod_{k=1}^{n}{cosleft( frac{t}{2^k} right)} = frac{sin t}{2^n sinleft(frac{t}{2^n} right)}$?
– Noor AlYaqeen
33 mins ago













1














$defe{mathrm{e}}$Another method: Note that $a + b = dfrac{a^2 - b^2}{a - b}$ for $a ≠ b$, then$$
M_{Z_n}(t) = frac{1}{2^n} prod_{k = 1}^n left( expleft( frac{t}{2^k} right) + expleft( -frac{t}{2^k} right) right) = frac{1}{2^n} · frac{e^t - e^{-t}}{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}.
$$

Because$$
lim_{n → ∞} frac{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}{dfrac{t}{2^n} - left( -dfrac{t}{2^n} right)} = (e^x)'bigr|_{x = 0} = 1,
$$

then $limlimits_{n → ∞} M_{Z_n}(t) = dfrac{1}{2t} (e^t - e^{-t})$ and the result follows.






share|cite|improve this answer





















  • Thanks this is the best answer!
    – Noor AlYaqeen
    9 mins ago
















1














$defe{mathrm{e}}$Another method: Note that $a + b = dfrac{a^2 - b^2}{a - b}$ for $a ≠ b$, then$$
M_{Z_n}(t) = frac{1}{2^n} prod_{k = 1}^n left( expleft( frac{t}{2^k} right) + expleft( -frac{t}{2^k} right) right) = frac{1}{2^n} · frac{e^t - e^{-t}}{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}.
$$

Because$$
lim_{n → ∞} frac{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}{dfrac{t}{2^n} - left( -dfrac{t}{2^n} right)} = (e^x)'bigr|_{x = 0} = 1,
$$

then $limlimits_{n → ∞} M_{Z_n}(t) = dfrac{1}{2t} (e^t - e^{-t})$ and the result follows.






share|cite|improve this answer





















  • Thanks this is the best answer!
    – Noor AlYaqeen
    9 mins ago














1












1








1






$defe{mathrm{e}}$Another method: Note that $a + b = dfrac{a^2 - b^2}{a - b}$ for $a ≠ b$, then$$
M_{Z_n}(t) = frac{1}{2^n} prod_{k = 1}^n left( expleft( frac{t}{2^k} right) + expleft( -frac{t}{2^k} right) right) = frac{1}{2^n} · frac{e^t - e^{-t}}{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}.
$$

Because$$
lim_{n → ∞} frac{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}{dfrac{t}{2^n} - left( -dfrac{t}{2^n} right)} = (e^x)'bigr|_{x = 0} = 1,
$$

then $limlimits_{n → ∞} M_{Z_n}(t) = dfrac{1}{2t} (e^t - e^{-t})$ and the result follows.






share|cite|improve this answer












$defe{mathrm{e}}$Another method: Note that $a + b = dfrac{a^2 - b^2}{a - b}$ for $a ≠ b$, then$$
M_{Z_n}(t) = frac{1}{2^n} prod_{k = 1}^n left( expleft( frac{t}{2^k} right) + expleft( -frac{t}{2^k} right) right) = frac{1}{2^n} · frac{e^t - e^{-t}}{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}.
$$

Because$$
lim_{n → ∞} frac{expleft( dfrac{t}{2^n} right) - expleft( -dfrac{t}{2^n} right)}{dfrac{t}{2^n} - left( -dfrac{t}{2^n} right)} = (e^x)'bigr|_{x = 0} = 1,
$$

then $limlimits_{n → ∞} M_{Z_n}(t) = dfrac{1}{2t} (e^t - e^{-t})$ and the result follows.







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answered 2 hours ago









Saad

19.7k92252




19.7k92252












  • Thanks this is the best answer!
    – Noor AlYaqeen
    9 mins ago


















  • Thanks this is the best answer!
    – Noor AlYaqeen
    9 mins ago
















Thanks this is the best answer!
– Noor AlYaqeen
9 mins ago




Thanks this is the best answer!
– Noor AlYaqeen
9 mins ago











1














In addition to characteristic functions, one may also approach the problem via binary representation of integers, which is not as short as the answers with characteristic functions, but is quite straightforward.



Indeed, rewrite
$$
Z_n = frac{1}{2^n} sumlimits_{j=1}^n 2^{n-j} X_j : = frac{1}{2^n}S_n.
$$

Define $Lambda_{+} = { 1leq j leq 2^n: X_j = 1 }$, and let $Lambda_-$ be the complement of $Lambda_+$ in $1leq jleq 2^n$. Then,
$$
S_n = sum_{jin Lambda_+} - sum_{j in Lambda_-} = sum_{jin Lambda_+} - left( 2^n - 1 - sumlimits_{jin Lambda_+} right) = 2sumlimits_{j in Lambda_+} 2^{n-j} - (2^n - 1) tag{1}.
$$

Thus, with $S_n$ we cover all integers from $-(2^n - 1) , ... 2^n - 1$ of the form $(1)$, which are precisely all the odd integers from $-(2^n - 1), ... (2^n - 1)$, $2^n$ in total. Thus, if $i in [-(2^n - 1), ..., 2^n - 1] $ is even then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 0 tag{2}
$$

and if $i $ is odd, then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 2^{-n}, tag{3}
$$

since there is a single choice of index set $Lambda_+$ in $(1)$, and hence $(3)$ follows in view of independence of ${X_j}$.



From $(2)$ and $(3)$ we see, by counting the number of odd integers, that for any integer $-2^{n-1} + 1 leq i leq 2^{n-1}$ one has
$$
mathbb{P}left(Z_n leq frac{2i - 1}{2^n} right) = frac{1}{2} + frac{i}{2^n}.
$$



It follows that the distribution function $F_n$ of $Z_n$ coincides, on odd dyadic rationals from $[-1,1]$ (and obviously everywhere on $(-infty, -1] cup [1,infty)$) with the distribution function $F$ of a random variable with the law $U[-1, 1]$. The density of dyadic rationals and right-continuity of cdf imply $F_n to F$ everywhere on $[-1,1]$, hence the claim.






share|cite|improve this answer





















  • Thank you, a very interesting solution! although, a bit complicated.
    – Noor AlYaqeen
    21 mins ago
















1














In addition to characteristic functions, one may also approach the problem via binary representation of integers, which is not as short as the answers with characteristic functions, but is quite straightforward.



Indeed, rewrite
$$
Z_n = frac{1}{2^n} sumlimits_{j=1}^n 2^{n-j} X_j : = frac{1}{2^n}S_n.
$$

Define $Lambda_{+} = { 1leq j leq 2^n: X_j = 1 }$, and let $Lambda_-$ be the complement of $Lambda_+$ in $1leq jleq 2^n$. Then,
$$
S_n = sum_{jin Lambda_+} - sum_{j in Lambda_-} = sum_{jin Lambda_+} - left( 2^n - 1 - sumlimits_{jin Lambda_+} right) = 2sumlimits_{j in Lambda_+} 2^{n-j} - (2^n - 1) tag{1}.
$$

Thus, with $S_n$ we cover all integers from $-(2^n - 1) , ... 2^n - 1$ of the form $(1)$, which are precisely all the odd integers from $-(2^n - 1), ... (2^n - 1)$, $2^n$ in total. Thus, if $i in [-(2^n - 1), ..., 2^n - 1] $ is even then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 0 tag{2}
$$

and if $i $ is odd, then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 2^{-n}, tag{3}
$$

since there is a single choice of index set $Lambda_+$ in $(1)$, and hence $(3)$ follows in view of independence of ${X_j}$.



From $(2)$ and $(3)$ we see, by counting the number of odd integers, that for any integer $-2^{n-1} + 1 leq i leq 2^{n-1}$ one has
$$
mathbb{P}left(Z_n leq frac{2i - 1}{2^n} right) = frac{1}{2} + frac{i}{2^n}.
$$



It follows that the distribution function $F_n$ of $Z_n$ coincides, on odd dyadic rationals from $[-1,1]$ (and obviously everywhere on $(-infty, -1] cup [1,infty)$) with the distribution function $F$ of a random variable with the law $U[-1, 1]$. The density of dyadic rationals and right-continuity of cdf imply $F_n to F$ everywhere on $[-1,1]$, hence the claim.






share|cite|improve this answer





















  • Thank you, a very interesting solution! although, a bit complicated.
    – Noor AlYaqeen
    21 mins ago














1












1








1






In addition to characteristic functions, one may also approach the problem via binary representation of integers, which is not as short as the answers with characteristic functions, but is quite straightforward.



Indeed, rewrite
$$
Z_n = frac{1}{2^n} sumlimits_{j=1}^n 2^{n-j} X_j : = frac{1}{2^n}S_n.
$$

Define $Lambda_{+} = { 1leq j leq 2^n: X_j = 1 }$, and let $Lambda_-$ be the complement of $Lambda_+$ in $1leq jleq 2^n$. Then,
$$
S_n = sum_{jin Lambda_+} - sum_{j in Lambda_-} = sum_{jin Lambda_+} - left( 2^n - 1 - sumlimits_{jin Lambda_+} right) = 2sumlimits_{j in Lambda_+} 2^{n-j} - (2^n - 1) tag{1}.
$$

Thus, with $S_n$ we cover all integers from $-(2^n - 1) , ... 2^n - 1$ of the form $(1)$, which are precisely all the odd integers from $-(2^n - 1), ... (2^n - 1)$, $2^n$ in total. Thus, if $i in [-(2^n - 1), ..., 2^n - 1] $ is even then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 0 tag{2}
$$

and if $i $ is odd, then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 2^{-n}, tag{3}
$$

since there is a single choice of index set $Lambda_+$ in $(1)$, and hence $(3)$ follows in view of independence of ${X_j}$.



From $(2)$ and $(3)$ we see, by counting the number of odd integers, that for any integer $-2^{n-1} + 1 leq i leq 2^{n-1}$ one has
$$
mathbb{P}left(Z_n leq frac{2i - 1}{2^n} right) = frac{1}{2} + frac{i}{2^n}.
$$



It follows that the distribution function $F_n$ of $Z_n$ coincides, on odd dyadic rationals from $[-1,1]$ (and obviously everywhere on $(-infty, -1] cup [1,infty)$) with the distribution function $F$ of a random variable with the law $U[-1, 1]$. The density of dyadic rationals and right-continuity of cdf imply $F_n to F$ everywhere on $[-1,1]$, hence the claim.






share|cite|improve this answer












In addition to characteristic functions, one may also approach the problem via binary representation of integers, which is not as short as the answers with characteristic functions, but is quite straightforward.



Indeed, rewrite
$$
Z_n = frac{1}{2^n} sumlimits_{j=1}^n 2^{n-j} X_j : = frac{1}{2^n}S_n.
$$

Define $Lambda_{+} = { 1leq j leq 2^n: X_j = 1 }$, and let $Lambda_-$ be the complement of $Lambda_+$ in $1leq jleq 2^n$. Then,
$$
S_n = sum_{jin Lambda_+} - sum_{j in Lambda_-} = sum_{jin Lambda_+} - left( 2^n - 1 - sumlimits_{jin Lambda_+} right) = 2sumlimits_{j in Lambda_+} 2^{n-j} - (2^n - 1) tag{1}.
$$

Thus, with $S_n$ we cover all integers from $-(2^n - 1) , ... 2^n - 1$ of the form $(1)$, which are precisely all the odd integers from $-(2^n - 1), ... (2^n - 1)$, $2^n$ in total. Thus, if $i in [-(2^n - 1), ..., 2^n - 1] $ is even then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 0 tag{2}
$$

and if $i $ is odd, then
$$
mathbb{P}left(Z_n = frac{i}{2^n}right) = 2^{-n}, tag{3}
$$

since there is a single choice of index set $Lambda_+$ in $(1)$, and hence $(3)$ follows in view of independence of ${X_j}$.



From $(2)$ and $(3)$ we see, by counting the number of odd integers, that for any integer $-2^{n-1} + 1 leq i leq 2^{n-1}$ one has
$$
mathbb{P}left(Z_n leq frac{2i - 1}{2^n} right) = frac{1}{2} + frac{i}{2^n}.
$$



It follows that the distribution function $F_n$ of $Z_n$ coincides, on odd dyadic rationals from $[-1,1]$ (and obviously everywhere on $(-infty, -1] cup [1,infty)$) with the distribution function $F$ of a random variable with the law $U[-1, 1]$. The density of dyadic rationals and right-continuity of cdf imply $F_n to F$ everywhere on $[-1,1]$, hence the claim.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 1 hour ago









Hayk

2,042213




2,042213












  • Thank you, a very interesting solution! although, a bit complicated.
    – Noor AlYaqeen
    21 mins ago


















  • Thank you, a very interesting solution! although, a bit complicated.
    – Noor AlYaqeen
    21 mins ago
















Thank you, a very interesting solution! although, a bit complicated.
– Noor AlYaqeen
21 mins ago




Thank you, a very interesting solution! although, a bit complicated.
– Noor AlYaqeen
21 mins ago


















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