Limiting a sequence of moment generating functions
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
add a comment |
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
Why not using characteristic functions?
– Saad
Dec 24 '18 at 15:36
@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
Dec 24 '18 at 15:49
add a comment |
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
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
probability-theory probability-limit-theorems
edited Dec 24 '18 at 16:11
Saad
19.7k92352
19.7k92352
asked Dec 24 '18 at 15:02
Noor AlYaqeenNoor AlYaqeen
346
346
Why not using characteristic functions?
– Saad
Dec 24 '18 at 15:36
@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
Dec 24 '18 at 15:49
add a comment |
Why not using characteristic functions?
– Saad
Dec 24 '18 at 15:36
@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
Dec 24 '18 at 15:49
Why not using characteristic functions?
– Saad
Dec 24 '18 at 15:36
Why not using characteristic functions?
– Saad
Dec 24 '18 at 15:36
@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
Dec 24 '18 at 15:49
@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
Dec 24 '18 at 15:49
add a comment |
3 Answers
3
active
oldest
votes
$defe{mathrm{e}}$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.
Thanks this is the best answer!
– Noor AlYaqeen
Dec 24 '18 at 18:45
add a comment |
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.
Thank you, a very interesting solution! although, a bit complicated.
– Noor AlYaqeen
Dec 24 '18 at 18:33
add a comment |
$defi{mathrm{i}}defd{mathrm{d}}$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, $Z_n xrightarrow{mathrm{d}} Z$.
I didn't know you could give more than one answer...
– maridia
Dec 24 '18 at 17:00
@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
Dec 24 '18 at 18:21
@maridia: After you've posted an answer, the "Your Answer" textbox goes away, but you get an "Add Another Answer" button in its place. If you click that button, you get a confirmation dialog, "Are you sure you want to add another answer? ¶ You could use the edit link to refine and improve your existing answer, instead." If you click 'OK', you get the "Your Answer" textbox back.
– ruakh
Dec 24 '18 at 21:32
@NoorAlYaqeen Note that $cosθ=dfrac{sin2θ}{sinθ}$.
– Saad
Dec 25 '18 at 0:02
add a comment |
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3 Answers
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3 Answers
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$defe{mathrm{e}}$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.
Thanks this is the best answer!
– Noor AlYaqeen
Dec 24 '18 at 18:45
add a comment |
$defe{mathrm{e}}$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.
Thanks this is the best answer!
– Noor AlYaqeen
Dec 24 '18 at 18:45
add a comment |
$defe{mathrm{e}}$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.
$defe{mathrm{e}}$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.
answered Dec 24 '18 at 16:30
SaadSaad
19.7k92352
19.7k92352
Thanks this is the best answer!
– Noor AlYaqeen
Dec 24 '18 at 18:45
add a comment |
Thanks this is the best answer!
– Noor AlYaqeen
Dec 24 '18 at 18:45
Thanks this is the best answer!
– Noor AlYaqeen
Dec 24 '18 at 18:45
Thanks this is the best answer!
– Noor AlYaqeen
Dec 24 '18 at 18:45
add a comment |
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.
Thank you, a very interesting solution! although, a bit complicated.
– Noor AlYaqeen
Dec 24 '18 at 18:33
add a comment |
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.
Thank you, a very interesting solution! although, a bit complicated.
– Noor AlYaqeen
Dec 24 '18 at 18:33
add a comment |
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.
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.
answered Dec 24 '18 at 17:04
HaykHayk
2,0721213
2,0721213
Thank you, a very interesting solution! although, a bit complicated.
– Noor AlYaqeen
Dec 24 '18 at 18:33
add a comment |
Thank you, a very interesting solution! although, a bit complicated.
– Noor AlYaqeen
Dec 24 '18 at 18:33
Thank you, a very interesting solution! although, a bit complicated.
– Noor AlYaqeen
Dec 24 '18 at 18:33
Thank you, a very interesting solution! although, a bit complicated.
– Noor AlYaqeen
Dec 24 '18 at 18:33
add a comment |
$defi{mathrm{i}}defd{mathrm{d}}$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, $Z_n xrightarrow{mathrm{d}} Z$.
I didn't know you could give more than one answer...
– maridia
Dec 24 '18 at 17:00
@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
Dec 24 '18 at 18:21
@maridia: After you've posted an answer, the "Your Answer" textbox goes away, but you get an "Add Another Answer" button in its place. If you click that button, you get a confirmation dialog, "Are you sure you want to add another answer? ¶ You could use the edit link to refine and improve your existing answer, instead." If you click 'OK', you get the "Your Answer" textbox back.
– ruakh
Dec 24 '18 at 21:32
@NoorAlYaqeen Note that $cosθ=dfrac{sin2θ}{sinθ}$.
– Saad
Dec 25 '18 at 0:02
add a comment |
$defi{mathrm{i}}defd{mathrm{d}}$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, $Z_n xrightarrow{mathrm{d}} Z$.
I didn't know you could give more than one answer...
– maridia
Dec 24 '18 at 17:00
@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
Dec 24 '18 at 18:21
@maridia: After you've posted an answer, the "Your Answer" textbox goes away, but you get an "Add Another Answer" button in its place. If you click that button, you get a confirmation dialog, "Are you sure you want to add another answer? ¶ You could use the edit link to refine and improve your existing answer, instead." If you click 'OK', you get the "Your Answer" textbox back.
– ruakh
Dec 24 '18 at 21:32
@NoorAlYaqeen Note that $cosθ=dfrac{sin2θ}{sinθ}$.
– Saad
Dec 25 '18 at 0:02
add a comment |
$defi{mathrm{i}}defd{mathrm{d}}$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, $Z_n xrightarrow{mathrm{d}} Z$.
$defi{mathrm{i}}defd{mathrm{d}}$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, $Z_n xrightarrow{mathrm{d}} Z$.
answered Dec 24 '18 at 16:08
SaadSaad
19.7k92352
19.7k92352
I didn't know you could give more than one answer...
– maridia
Dec 24 '18 at 17:00
@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
Dec 24 '18 at 18:21
@maridia: After you've posted an answer, the "Your Answer" textbox goes away, but you get an "Add Another Answer" button in its place. If you click that button, you get a confirmation dialog, "Are you sure you want to add another answer? ¶ You could use the edit link to refine and improve your existing answer, instead." If you click 'OK', you get the "Your Answer" textbox back.
– ruakh
Dec 24 '18 at 21:32
@NoorAlYaqeen Note that $cosθ=dfrac{sin2θ}{sinθ}$.
– Saad
Dec 25 '18 at 0:02
add a comment |
I didn't know you could give more than one answer...
– maridia
Dec 24 '18 at 17:00
@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
Dec 24 '18 at 18:21
@maridia: After you've posted an answer, the "Your Answer" textbox goes away, but you get an "Add Another Answer" button in its place. If you click that button, you get a confirmation dialog, "Are you sure you want to add another answer? ¶ You could use the edit link to refine and improve your existing answer, instead." If you click 'OK', you get the "Your Answer" textbox back.
– ruakh
Dec 24 '18 at 21:32
@NoorAlYaqeen Note that $cosθ=dfrac{sin2θ}{sinθ}$.
– Saad
Dec 25 '18 at 0:02
I didn't know you could give more than one answer...
– maridia
Dec 24 '18 at 17:00
I didn't know you could give more than one answer...
– maridia
Dec 24 '18 at 17:00
@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
Dec 24 '18 at 18:21
@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
Dec 24 '18 at 18:21
@maridia: After you've posted an answer, the "Your Answer" textbox goes away, but you get an "Add Another Answer" button in its place. If you click that button, you get a confirmation dialog, "Are you sure you want to add another answer? ¶ You could use the edit link to refine and improve your existing answer, instead." If you click 'OK', you get the "Your Answer" textbox back.
– ruakh
Dec 24 '18 at 21:32
@maridia: After you've posted an answer, the "Your Answer" textbox goes away, but you get an "Add Another Answer" button in its place. If you click that button, you get a confirmation dialog, "Are you sure you want to add another answer? ¶ You could use the edit link to refine and improve your existing answer, instead." If you click 'OK', you get the "Your Answer" textbox back.
– ruakh
Dec 24 '18 at 21:32
@NoorAlYaqeen Note that $cosθ=dfrac{sin2θ}{sinθ}$.
– Saad
Dec 25 '18 at 0:02
@NoorAlYaqeen Note that $cosθ=dfrac{sin2θ}{sinθ}$.
– Saad
Dec 25 '18 at 0:02
add a comment |
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Why not using characteristic functions?
– Saad
Dec 24 '18 at 15:36
@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
Dec 24 '18 at 15:49