Derivation for marginal effect sizes' distribution in linear model
$begingroup$
Model:
$$Y_{Ntimes 1} = X_{Ntimes M} beta_{Mtimes 1} + epsilon_{Ntimes 1}$$
the design matrix $X_{Ntimes M}$ is known, each column of the design matrix has been standardized to have mean 0 and variance 1. $Y$ is also known and has been standardized to mean 0 and variance 1. Also assume $beta sim mathcal{N}(0, sigma^2 I_{Mtimes M})$ and $epsilon sim mathcal{N}(0, sigma_epsilon^2 I_{Ntimes N})$. In addition $M sigma^2 + sigma_epsilon^2 = 1$
The marginal effect sizes are
$hat{beta} = frac{1}{N}X^top Y$. The distribution the marginal effect sizes $hat{beta}$ can be determined as follows:
begin{align*}
hat{beta} & = frac{X^top Y}{N} \
& = frac{X^top ( X beta + epsilon) }{N} \
& = frac{XX^top}{N} beta + frac{1}{N}X^top epsilon
end{align*}
$$E[hat{beta} | X, beta] = frac{XX^top}{N} beta$$
$$text{Var}[hat{beta} | X,beta] = frac{1}{N^2}X^top E[epsilon epsilon^top]X=frac{1}{N^2}sigma_epsilon^2X^top X$$
Thus $$hat{beta} sim mathcal{N}(frac{XX^top}{N} beta, frac{1}{N^2}sigma_epsilon^2X^top X)$$
My question is that if we consider a small chunk of variable of size b, where the corresponding design matrix is of shape $X_{N times b} $, the corresponding marginal effect sizes is $hat{beta}_b$:
$$hat{beta}_b = frac{X^top_b Y}{N}$$
How is the following two derived:
$$E[hat{beta}_b | beta_b, X_b] = frac{X_b^top X_b}{N} beta_b$$
$$text{Var}[hat{beta}_b | beta_b, X_b] = frac{1}{N}( 1 - b sigma^2) frac{X_b^top X_b}{N}$$
One thing I am confused about is the $X_b$ being conditioned in $E[hat{beta}_b | beta_b, X_b]$. Does this mean that we don't assume anything on the rest of the design matrix?
For more context, this is from the methods section in this paper.
statistical-inference linear-regression
asked Jan 14 at 15:07
DavidDavid
286
$endgroup$
add a comment |
$begingroup$
Model:
$$Y_{Ntimes 1} = X_{Ntimes M} beta_{Mtimes 1} + epsilon_{Ntimes 1}$$
the design matrix $X_{Ntimes M}$ is known, each column of the design matrix has been standardized to have mean 0 and variance 1. $Y$ is also known and has been standardized to mean 0 and variance 1. Also assume $beta sim mathcal{N}(0, sigma^2 I_{Mtimes M})$ and $epsilon sim mathcal{N}(0, sigma_epsilon^2 I_{Ntimes N})$. In addition $M sigma^2 + sigma_epsilon^2 = 1$
The marginal effect sizes are
$hat{beta} = frac{1}{N}X^top Y$. The distribution the marginal effect sizes $hat{beta}$ can be determined as follows:
begin{align*}
hat{beta} & = frac{X^top Y}{N} \
& = frac{X^top ( X beta + epsilon) }{N} \
& = frac{XX^top}{N} beta + frac{1}{N}X^top epsilon
end{align*}
$$E[hat{beta} | X, beta] = frac{XX^top}{N} beta$$
$$text{Var}[hat{beta} | X,beta] = frac{1}{N^2}X^top E[epsilon epsilon^top]X=frac{1}{N^2}sigma_epsilon^2X^top X$$
Thus $$hat{beta} sim mathcal{N}(frac{XX^top}{N} beta, frac{1}{N^2}sigma_epsilon^2X^top X)$$
My question is that if we consider a small chunk of variable of size b, where the corresponding design matrix is of shape $X_{N times b} $, the corresponding marginal effect sizes is $hat{beta}_b$:
$$hat{beta}_b = frac{X^top_b Y}{N}$$
How is the following two derived:
$$E[hat{beta}_b | beta_b, X_b] = frac{X_b^top X_b}{N} beta_b$$
$$text{Var}[hat{beta}_b | beta_b, X_b] = frac{1}{N}( 1 - b sigma^2) frac{X_b^top X_b}{N}$$
One thing I am confused about is the $X_b$ being conditioned in $E[hat{beta}_b | beta_b, X_b]$. Does this mean that we don't assume anything on the rest of the design matrix?
For more context, this is from the methods section in this paper.
statistical-inference linear-regression
asked Jan 14 at 15:07
DavidDavid
286
$endgroup$
add a comment |
$begingroup$
Model:
$$Y_{Ntimes 1} = X_{Ntimes M} beta_{Mtimes 1} + epsilon_{Ntimes 1}$$
the design matrix $X_{Ntimes M}$ is known, each column of the design matrix has been standardized to have mean 0 and variance 1. $Y$ is also known and has been standardized to mean 0 and variance 1. Also assume $beta sim mathcal{N}(0, sigma^2 I_{Mtimes M})$ and $epsilon sim mathcal{N}(0, sigma_epsilon^2 I_{Ntimes N})$. In addition $M sigma^2 + sigma_epsilon^2 = 1$
The marginal effect sizes are
$hat{beta} = frac{1}{N}X^top Y$. The distribution the marginal effect sizes $hat{beta}$ can be determined as follows:
begin{align*}
hat{beta} & = frac{X^top Y}{N} \
& = frac{X^top ( X beta + epsilon) }{N} \
& = frac{XX^top}{N} beta + frac{1}{N}X^top epsilon
end{align*}
$$E[hat{beta} | X, beta] = frac{XX^top}{N} beta$$
$$text{Var}[hat{beta} | X,beta] = frac{1}{N^2}X^top E[epsilon epsilon^top]X=frac{1}{N^2}sigma_epsilon^2X^top X$$
Thus $$hat{beta} sim mathcal{N}(frac{XX^top}{N} beta, frac{1}{N^2}sigma_epsilon^2X^top X)$$
My question is that if we consider a small chunk of variable of size b, where the corresponding design matrix is of shape $X_{N times b} $, the corresponding marginal effect sizes is $hat{beta}_b$:
$$hat{beta}_b = frac{X^top_b Y}{N}$$
How is the following two derived:
$$E[hat{beta}_b | beta_b, X_b] = frac{X_b^top X_b}{N} beta_b$$
$$text{Var}[hat{beta}_b | beta_b, X_b] = frac{1}{N}( 1 - b sigma^2) frac{X_b^top X_b}{N}$$
One thing I am confused about is the $X_b$ being conditioned in $E[hat{beta}_b | beta_b, X_b]$. Does this mean that we don't assume anything on the rest of the design matrix?
For more context, this is from the methods section in this paper.
statistical-inference linear-regression
asked Jan 14 at 15:07
DavidDavid
286
$endgroup$
Model:
$$Y_{Ntimes 1} = X_{Ntimes M} beta_{Mtimes 1} + epsilon_{Ntimes 1}$$
the design matrix $X_{Ntimes M}$ is known, each column of the design matrix has been standardized to have mean 0 and variance 1. $Y$ is also known and has been standardized to mean 0 and variance 1. Also assume $beta sim mathcal{N}(0, sigma^2 I_{Mtimes M})$ and $epsilon sim mathcal{N}(0, sigma_epsilon^2 I_{Ntimes N})$. In addition $M sigma^2 + sigma_epsilon^2 = 1$
The marginal effect sizes are
$hat{beta} = frac{1}{N}X^top Y$. The distribution the marginal effect sizes $hat{beta}$ can be determined as follows:
begin{align*}
hat{beta} & = frac{X^top Y}{N} \
& = frac{X^top ( X beta + epsilon) }{N} \
& = frac{XX^top}{N} beta + frac{1}{N}X^top epsilon
end{align*}
$$E[hat{beta} | X, beta] = frac{XX^top}{N} beta$$
$$text{Var}[hat{beta} | X,beta] = frac{1}{N^2}X^top E[epsilon epsilon^top]X=frac{1}{N^2}sigma_epsilon^2X^top X$$
Thus $$hat{beta} sim mathcal{N}(frac{XX^top}{N} beta, frac{1}{N^2}sigma_epsilon^2X^top X)$$
My question is that if we consider a small chunk of variable of size b, where the corresponding design matrix is of shape $X_{N times b} $, the corresponding marginal effect sizes is $hat{beta}_b$:
$$hat{beta}_b = frac{X^top_b Y}{N}$$
How is the following two derived:
$$E[hat{beta}_b | beta_b, X_b] = frac{X_b^top X_b}{N} beta_b$$
$$text{Var}[hat{beta}_b | beta_b, X_b] = frac{1}{N}( 1 - b sigma^2) frac{X_b^top X_b}{N}$$
One thing I am confused about is the $X_b$ being conditioned in $E[hat{beta}_b | beta_b, X_b]$. Does this mean that we don't assume anything on the rest of the design matrix?
For more context, this is from the methods section in this paper.
statistical-inference linear-regression
statistical-inference linear-regression
asked Jan 14 at 15:07
DavidDavid
286
asked Jan 14 at 15:07
DavidDavid
286
asked Jan 14 at 15:07
DavidDavid
286
asked Jan 14 at 15:07
DavidDavid
286
asked Jan 14 at 15:07
DavidDavid
286
286
add a comment |
add a comment |
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