The maximum expected deviation from the sample average matrix?
$begingroup$
I have reached to $mathbb{E}[|x_tx_t^T - G_t|^2_F]$, and I need an upperbound for it in terms of probabilistic characteristics of $x_t$ where:
$x_t$ is a random vector in $mathbb{R}^n$ drawn from an i.i.d with mean of $mu$ and variance $sigma^2$
$G_t$ is a symmetric matrix which I want it to be the sample average of $x_tx_t^T$ (I am not sure whether it can be simply the average of all $x_tx_t^T$ matrices according to the following or not?).
To handle the stated problem, I am going to generalize vector problem to matrix one as the following:
Assume we want to find an upperbound for $mathbb{E}[|x_t - z_t|^2]$
$$
|x_t - z_t|^2=|x_t - mu + mu -z|^2 = |x_t - mu |^2 + |mu -z_t|^2 +
2langle x_t - mu , mu -z_trangle
$$
where $z_t = frac{1}{t-1}sum_{s=1}^{t-1}x_s$ is the average of $t-1$ number of data. Taking the expected value vanishes the last part and gives $sigma^2$ as the variance of data.
$$
mathbb{E}[|x_t - z_t|^2]=mathbb{E}[|x_t - mu |^2] + mathbb{E}[|z_t- mu |^2]=sigma^2 + mathbb{E}[|mu -z_t|^2]
$$
Also, $mathbb{E}[|z_t- mu |^2] = mathbb{E}[|frac{1}{t-1}x_1 -frac{1}{t-1}mu +cdots + frac{1}{t-1}x_{t-1} -frac{1}{t-1}mu|^2] leq frac{1}{(t-1)^2}sigma^2$
Therefore,
$$
mathbb{E}[|x_t - z_t|^2] leq sigma^2 + frac{1}{(t-1)^2}sigma^2
$$
Now, my questions are:
1- Is the sample average of $x_tx_t^T$ can be defined simply as $G_t=frac{1}{t-1}sum_{s=1}^{t-1}x_sx_s^T$ because $z_t$ had beautiful property that helped me to simplify the expression.
2- Is there any $G_t$ in the matrix form that I can use it to get some good result?
My guess is $G_t$ has to be something which gives us an upperbound in terms of covariance of the data:
$$|x_tx_t^T - G_t|^2_F$$
probability covariance variance expected-value
asked Dec 23 '18 at 4:34
SaeedSaeed
1,149310
$endgroup$
add a comment |
$begingroup$
I have reached to $mathbb{E}[|x_tx_t^T - G_t|^2_F]$, and I need an upperbound for it in terms of probabilistic characteristics of $x_t$ where:
$x_t$ is a random vector in $mathbb{R}^n$ drawn from an i.i.d with mean of $mu$ and variance $sigma^2$
$G_t$ is a symmetric matrix which I want it to be the sample average of $x_tx_t^T$ (I am not sure whether it can be simply the average of all $x_tx_t^T$ matrices according to the following or not?).
To handle the stated problem, I am going to generalize vector problem to matrix one as the following:
Assume we want to find an upperbound for $mathbb{E}[|x_t - z_t|^2]$
$$
|x_t - z_t|^2=|x_t - mu + mu -z|^2 = |x_t - mu |^2 + |mu -z_t|^2 +
2langle x_t - mu , mu -z_trangle
$$
where $z_t = frac{1}{t-1}sum_{s=1}^{t-1}x_s$ is the average of $t-1$ number of data. Taking the expected value vanishes the last part and gives $sigma^2$ as the variance of data.
$$
mathbb{E}[|x_t - z_t|^2]=mathbb{E}[|x_t - mu |^2] + mathbb{E}[|z_t- mu |^2]=sigma^2 + mathbb{E}[|mu -z_t|^2]
$$
Also, $mathbb{E}[|z_t- mu |^2] = mathbb{E}[|frac{1}{t-1}x_1 -frac{1}{t-1}mu +cdots + frac{1}{t-1}x_{t-1} -frac{1}{t-1}mu|^2] leq frac{1}{(t-1)^2}sigma^2$
Therefore,
$$
mathbb{E}[|x_t - z_t|^2] leq sigma^2 + frac{1}{(t-1)^2}sigma^2
$$
Now, my questions are:
1- Is the sample average of $x_tx_t^T$ can be defined simply as $G_t=frac{1}{t-1}sum_{s=1}^{t-1}x_sx_s^T$ because $z_t$ had beautiful property that helped me to simplify the expression.
2- Is there any $G_t$ in the matrix form that I can use it to get some good result?
My guess is $G_t$ has to be something which gives us an upperbound in terms of covariance of the data:
$$|x_tx_t^T - G_t|^2_F$$
probability covariance variance expected-value
asked Dec 23 '18 at 4:34
SaeedSaeed
1,149310
$endgroup$
add a comment |
$begingroup$
I have reached to $mathbb{E}[|x_tx_t^T - G_t|^2_F]$, and I need an upperbound for it in terms of probabilistic characteristics of $x_t$ where:
$x_t$ is a random vector in $mathbb{R}^n$ drawn from an i.i.d with mean of $mu$ and variance $sigma^2$
$G_t$ is a symmetric matrix which I want it to be the sample average of $x_tx_t^T$ (I am not sure whether it can be simply the average of all $x_tx_t^T$ matrices according to the following or not?).
To handle the stated problem, I am going to generalize vector problem to matrix one as the following:
Assume we want to find an upperbound for $mathbb{E}[|x_t - z_t|^2]$
$$
|x_t - z_t|^2=|x_t - mu + mu -z|^2 = |x_t - mu |^2 + |mu -z_t|^2 +
2langle x_t - mu , mu -z_trangle
$$
where $z_t = frac{1}{t-1}sum_{s=1}^{t-1}x_s$ is the average of $t-1$ number of data. Taking the expected value vanishes the last part and gives $sigma^2$ as the variance of data.
$$
mathbb{E}[|x_t - z_t|^2]=mathbb{E}[|x_t - mu |^2] + mathbb{E}[|z_t- mu |^2]=sigma^2 + mathbb{E}[|mu -z_t|^2]
$$
Also, $mathbb{E}[|z_t- mu |^2] = mathbb{E}[|frac{1}{t-1}x_1 -frac{1}{t-1}mu +cdots + frac{1}{t-1}x_{t-1} -frac{1}{t-1}mu|^2] leq frac{1}{(t-1)^2}sigma^2$
Therefore,
$$
mathbb{E}[|x_t - z_t|^2] leq sigma^2 + frac{1}{(t-1)^2}sigma^2
$$
Now, my questions are:
1- Is the sample average of $x_tx_t^T$ can be defined simply as $G_t=frac{1}{t-1}sum_{s=1}^{t-1}x_sx_s^T$ because $z_t$ had beautiful property that helped me to simplify the expression.
2- Is there any $G_t$ in the matrix form that I can use it to get some good result?
My guess is $G_t$ has to be something which gives us an upperbound in terms of covariance of the data:
$$|x_tx_t^T - G_t|^2_F$$
probability covariance variance expected-value
asked Dec 23 '18 at 4:34
SaeedSaeed
1,149310
$endgroup$
I have reached to $mathbb{E}[|x_tx_t^T - G_t|^2_F]$, and I need an upperbound for it in terms of probabilistic characteristics of $x_t$ where:
$x_t$ is a random vector in $mathbb{R}^n$ drawn from an i.i.d with mean of $mu$ and variance $sigma^2$
$G_t$ is a symmetric matrix which I want it to be the sample average of $x_tx_t^T$ (I am not sure whether it can be simply the average of all $x_tx_t^T$ matrices according to the following or not?).
To handle the stated problem, I am going to generalize vector problem to matrix one as the following:
Assume we want to find an upperbound for $mathbb{E}[|x_t - z_t|^2]$
$$
|x_t - z_t|^2=|x_t - mu + mu -z|^2 = |x_t - mu |^2 + |mu -z_t|^2 +
2langle x_t - mu , mu -z_trangle
$$
where $z_t = frac{1}{t-1}sum_{s=1}^{t-1}x_s$ is the average of $t-1$ number of data. Taking the expected value vanishes the last part and gives $sigma^2$ as the variance of data.
$$
mathbb{E}[|x_t - z_t|^2]=mathbb{E}[|x_t - mu |^2] + mathbb{E}[|z_t- mu |^2]=sigma^2 + mathbb{E}[|mu -z_t|^2]
$$
Also, $mathbb{E}[|z_t- mu |^2] = mathbb{E}[|frac{1}{t-1}x_1 -frac{1}{t-1}mu +cdots + frac{1}{t-1}x_{t-1} -frac{1}{t-1}mu|^2] leq frac{1}{(t-1)^2}sigma^2$
Therefore,
$$
mathbb{E}[|x_t - z_t|^2] leq sigma^2 + frac{1}{(t-1)^2}sigma^2
$$
Now, my questions are:
1- Is the sample average of $x_tx_t^T$ can be defined simply as $G_t=frac{1}{t-1}sum_{s=1}^{t-1}x_sx_s^T$ because $z_t$ had beautiful property that helped me to simplify the expression.
2- Is there any $G_t$ in the matrix form that I can use it to get some good result?
My guess is $G_t$ has to be something which gives us an upperbound in terms of covariance of the data:
$$|x_tx_t^T - G_t|^2_F$$
probability covariance variance expected-value
probability covariance variance expected-value
asked Dec 23 '18 at 4:34
SaeedSaeed
1,149310
asked Dec 23 '18 at 4:34
SaeedSaeed
1,149310
asked Dec 23 '18 at 4:34
SaeedSaeed
1,149310
asked Dec 23 '18 at 4:34
SaeedSaeed
1,149310
asked Dec 23 '18 at 4:34
SaeedSaeed
1,149310
1,149310
add a comment |
add a comment |
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