Show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant $2p$
Let $X_n$ ~ Bernoulli(p). Let $Y_n = X_n + X_{n+1}$.
Let $T_n = frac{1}{n}sum_{i=1}^{n} Y_i$.
I want to show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant 2p.
I found that $E[T_n] = 2p$ and that $operatorname{Var}[T_n] = 2p(1-p)frac{2n-1}{n^2}$.
My definition of convergence in probability is the following:
$$forall epsilon > 0 spacemathbb{P}(vert T_n - 2p vert > epsilon) to 0$$
I can also use the following criterion:
Convergence in probability iff $$lim_{ntoinfty} mathbb{E}Big[frac{vert T_n - 2pvert}{vert T_n - 2pvert + 1}Big] = 0$$
To me using the criterion here seems smart because I already know that the expected value is $2p$, but I am not sure how to proceed. Any hints?
probability convergence
edited Jan 4 at 21:57
Davide Giraudo
125k16150261
asked Jan 4 at 20:30
qcc101qcc101
499113
add a comment |
Let $X_n$ ~ Bernoulli(p). Let $Y_n = X_n + X_{n+1}$.
Let $T_n = frac{1}{n}sum_{i=1}^{n} Y_i$.
I want to show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant 2p.
I found that $E[T_n] = 2p$ and that $operatorname{Var}[T_n] = 2p(1-p)frac{2n-1}{n^2}$.
My definition of convergence in probability is the following:
$$forall epsilon > 0 spacemathbb{P}(vert T_n - 2p vert > epsilon) to 0$$
I can also use the following criterion:
Convergence in probability iff $$lim_{ntoinfty} mathbb{E}Big[frac{vert T_n - 2pvert}{vert T_n - 2pvert + 1}Big] = 0$$
To me using the criterion here seems smart because I already know that the expected value is $2p$, but I am not sure how to proceed. Any hints?
probability convergence
edited Jan 4 at 21:57
Davide Giraudo
125k16150261
asked Jan 4 at 20:30
qcc101qcc101
499113
Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
– dem0nakos
Jan 4 at 20:38
Can you use the law of large numbers?
– Lundborg
Jan 4 at 20:54
@dem0nakos I didn't know that inequality, super useful.
– qcc101
Jan 4 at 20:58
add a comment |
Let $X_n$ ~ Bernoulli(p). Let $Y_n = X_n + X_{n+1}$.
Let $T_n = frac{1}{n}sum_{i=1}^{n} Y_i$.
I want to show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant 2p.
I found that $E[T_n] = 2p$ and that $operatorname{Var}[T_n] = 2p(1-p)frac{2n-1}{n^2}$.
My definition of convergence in probability is the following:
$$forall epsilon > 0 spacemathbb{P}(vert T_n - 2p vert > epsilon) to 0$$
I can also use the following criterion:
Convergence in probability iff $$lim_{ntoinfty} mathbb{E}Big[frac{vert T_n - 2pvert}{vert T_n - 2pvert + 1}Big] = 0$$
To me using the criterion here seems smart because I already know that the expected value is $2p$, but I am not sure how to proceed. Any hints?
probability convergence
edited Jan 4 at 21:57
Davide Giraudo
125k16150261
asked Jan 4 at 20:30
qcc101qcc101
499113
Let $X_n$ ~ Bernoulli(p). Let $Y_n = X_n + X_{n+1}$.
Let $T_n = frac{1}{n}sum_{i=1}^{n} Y_i$.
I want to show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant 2p.
I found that $E[T_n] = 2p$ and that $operatorname{Var}[T_n] = 2p(1-p)frac{2n-1}{n^2}$.
My definition of convergence in probability is the following:
$$forall epsilon > 0 spacemathbb{P}(vert T_n - 2p vert > epsilon) to 0$$
I can also use the following criterion:
Convergence in probability iff $$lim_{ntoinfty} mathbb{E}Big[frac{vert T_n - 2pvert}{vert T_n - 2pvert + 1}Big] = 0$$
To me using the criterion here seems smart because I already know that the expected value is $2p$, but I am not sure how to proceed. Any hints?
probability convergence
probability convergence
edited Jan 4 at 21:57
Davide Giraudo
125k16150261
asked Jan 4 at 20:30
qcc101qcc101
499113
edited Jan 4 at 21:57
Davide Giraudo
125k16150261
asked Jan 4 at 20:30
qcc101qcc101
499113
edited Jan 4 at 21:57
Davide Giraudo
125k16150261
edited Jan 4 at 21:57
Davide Giraudo
125k16150261
edited Jan 4 at 21:57
Davide Giraudo
125k16150261
125k16150261
asked Jan 4 at 20:30
qcc101qcc101
499113
asked Jan 4 at 20:30
qcc101qcc101
499113
asked Jan 4 at 20:30
qcc101qcc101
499113
499113
Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
– dem0nakos
Jan 4 at 20:38
Can you use the law of large numbers?
– Lundborg
Jan 4 at 20:54
@dem0nakos I didn't know that inequality, super useful.
– qcc101
Jan 4 at 20:58
add a comment |
Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
– dem0nakos
Jan 4 at 20:38
Can you use the law of large numbers?
– Lundborg
Jan 4 at 20:54
@dem0nakos I didn't know that inequality, super useful.
– qcc101
Jan 4 at 20:58
Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
– dem0nakos
Jan 4 at 20:38
Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
– dem0nakos
Jan 4 at 20:38
Can you use the law of large numbers?
– Lundborg
Jan 4 at 20:54
Can you use the law of large numbers?
– Lundborg
Jan 4 at 20:54
@dem0nakos I didn't know that inequality, super useful.
– qcc101
Jan 4 at 20:58
@dem0nakos I didn't know that inequality, super useful.
– qcc101
Jan 4 at 20:58
add a comment |
2 Answers
2
active
oldest
votes
Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.
Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.
answered Jan 4 at 20:39
Will M.Will M.
2,442314
add a comment |
Note that
$$
T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
$$
Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that
$$
2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
$$
and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding
$$
T_n overset{P}{to} 2 p
$$
as desired.
answered Jan 4 at 20:58
LundborgLundborg
746414
add a comment |
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2 Answers
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2 Answers
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active
oldest
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active
oldest
votes
Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.
Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.
answered Jan 4 at 20:39
Will M.Will M.
2,442314
add a comment |
Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.
Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.
answered Jan 4 at 20:39
Will M.Will M.
2,442314
add a comment |
Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.
Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.
answered Jan 4 at 20:39
Will M.Will M.
2,442314
Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.
Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.
answered Jan 4 at 20:39
Will M.Will M.
2,442314
answered Jan 4 at 20:39
Will M.Will M.
2,442314
answered Jan 4 at 20:39
Will M.Will M.
2,442314
answered Jan 4 at 20:39
Will M.Will M.
2,442314
2,442314
add a comment |
add a comment |
Note that
$$
T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
$$
Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that
$$
2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
$$
and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding
$$
T_n overset{P}{to} 2 p
$$
as desired.
answered Jan 4 at 20:58
LundborgLundborg
746414
add a comment |
Note that
$$
T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
$$
Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that
$$
2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
$$
and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding
$$
T_n overset{P}{to} 2 p
$$
as desired.
answered Jan 4 at 20:58
LundborgLundborg
746414
add a comment |
Note that
$$
T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
$$
Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that
$$
2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
$$
and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding
$$
T_n overset{P}{to} 2 p
$$
as desired.
answered Jan 4 at 20:58
LundborgLundborg
746414
Note that
$$
T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
$$
Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that
$$
2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
$$
and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding
$$
T_n overset{P}{to} 2 p
$$
as desired.
answered Jan 4 at 20:58
LundborgLundborg
746414
answered Jan 4 at 20:58
LundborgLundborg
746414
answered Jan 4 at 20:58
LundborgLundborg
746414
answered Jan 4 at 20:58
LundborgLundborg
746414
746414
add a comment |
add a comment |
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Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
– dem0nakos
Jan 4 at 20:38
Can you use the law of large numbers?
– Lundborg
Jan 4 at 20:54
@dem0nakos I didn't know that inequality, super useful.
– qcc101
Jan 4 at 20:58