Kết hợp tuyến tính của hai biến ngẫu nhiên bình thường đa biến phụ thuộc

Giả sử chúng ta có hai vectơ của các biến ngẫu nhiên, cả hai đều bình thường, tức là $X \sim N(\mu_X, \Sigma_X)$ và $Y \sim N(\mu_Y, \Sigma_Y)$ . Chúng tôi quan tâm đến việc phân phối tổ hợp tuyến tính $Z = A X + B Y + C$ , trong đó $A$ và $B$ là ma trận, $C$ là một vectơ. Nếu $X$ và $Y$ là độc lập, . Câu hỏi là trong trường hợp phụ thuộc, giả sử rằng chúng ta biết mối tương quan của bất kỳ cặp nào . Cảm ơn bạn. $Z \sim N(A \mu_X + B \mu_Y + C, A \Sigma_X A^T + B \Sigma_Y B^T)$ $(X_i, Y_i)$

Lời chúc tốt đẹp nhất, Ivan

probability normal-distribution multinomial

— Ivan
nguồn

Câu trả lời:

Trong trường hợp đó, bạn phải ghi (với ký hiệu hy vọng rõ ràng) ( chỉnh sửa: giả sử bình thường chung của ) Sau đó và

(\begin{matrix} X \\ Y \end{matrix}) \sim N [(\begin{matrix} μ_{X} \\ μ_{Y} \end{matrix}), Σ_{X, Y}]

$\left(\begin{matrix}X\\Y \end{matrix}\right) \sim \mathcal{N}\left[ \left(\begin{matrix}\mu_X\\\mu_Y\end{matrix}\right), \Sigma_{X,Y} \right]$

(X, Y)

$(X,Y)$

A X + B Y = (\begin{matrix} A & B \end{matrix}) (\begin{matrix} X \\ Y \end{matrix})

$AX+BY=\left(\begin{matrix}A& B \end{matrix}\right) \left(\begin{matrix}X\\Y \end{matrix}\right)$

tức là

A X + B Y + C \sim N [(\begin{matrix} A & B \end{matrix}) (\begin{matrix} μ_{X} \\ μ_{Y} \end{matrix}) + C, (\begin{matrix} A & B \end{matrix}) Σ_{X, Y} (\begin{matrix} A^{T} \\ B^{T} \end{matrix})]

$AX+BY+C \sim \mathcal{N}\left[ \left(\begin{matrix}A& B \end{matrix}\right) \left(\begin{matrix}\mu_X\\\mu_Y\end{matrix}\right) + C, \left(\begin{matrix}A & B \end{matrix}\right)\Sigma_{X,Y} \left(\begin{matrix}A^T \\ B^T \end{matrix}\right)\right]$

A X + B Y + C \sim N [A μ_{X} + B μ_{Y} + C, A Σ_{X X} A^{T} + B Σ_{X Y}^{T} A^{T} + A Σ_{X Y} B^{T} + B Σ_{Y Y} B^{T}]

$AX+BY+C \sim \mathcal{N}\left[A\mu_X + B\mu_Y +C, A\Sigma_{XX}A^T+B\Sigma_{XY}^TA^T+A\Sigma_{XY}B^T+B\Sigma_{YY}B^T \right]$

— Tây An
nguồn

X

$X$

Y

$Y$

B Σ_{X Y}^{T} A^{T} + A Σ_{X Y} B^{T}

$B\Sigma_{XY}^TA^T+A\Sigma_{XY}B^T$

2 A Σ_{X Y} B^{T}

$2A\Sigma_{XY}B^T$

B Σ_{X Y}^{T} A^{T} + A Σ_{X Y} B^{T} = (A Σ_{X Y} B^{T})^{T} + A Σ_{X Y} B^{T}

$B\Sigma_{XY}^TA^T+A\Sigma_{XY}B^T = (A\Sigma_{XY}B^T)^T + A\Sigma_{XY}B^T$

Σ_{X Y}

$\Sigma_{XY}$

(i, j)

$(i,j)$

cov (X_{i}, Y_{j})

$\text{cov}(X_i,Y_j)$

(j, i)

$(j,i)$

cov (X_{j}, Y_{i})

$\text{cov}(X_j,Y_i)$ , and there is no reason why these covariances must be equal.

— Dilip Sarwate

@DilipSarwate: (+1) you are right, in the general case, there is no reason for these two terms to be equal.

— Xi'an

Your question does not have a unique answer as currently posed unless you assume that $X$ and $Y$ are jointly normally distributed with covariance top right block $\Sigma_{XY}$ . i think you do mean this because you say you have each covariance between X and Y. In this case we can write $W=(X^T,Y^T)^T$ which is also multivariate normal. then $Z$ is given in terms of $W$ as:

Z = (A, B) W + C

$Z=(A,B)W+C$

Then you use your usual formula for linear combination. Note that the mean is unchanged but the covariance matrix has two extra terms added $A\Sigma_{XY}B^T+B\Sigma_{XY}^TA^T$

— probabilityislogic
nguồn

Thanks for pointing out this issue, in fact, I didn't even think about it, but it seems that the variables are indeed can be viewed, in my case, as jointly normally distributed, even if their components are correlated.

— Ivan

I agree that the question cannot be solved as posed. It can be solved in a straightforward manner if one assumes, as @Xi'an's answer does, that

X

$X$ and

Y

$Y$ are jointly normally distributed. It could be solvable, presumably with more difficulty, if the joint distribution was specified as something other than joint normal. But just knowing

cov (X_{i}, Y_{j})

$\text{cov}(X_i,Y_j)$ for all

i, j

$i, j$ , does not mean that $W = (X^T,Y^T)^T$ is multivariate normal. Any two random variables with finite variances have a covariance. Covariance is not defined only for normal or jointly normal random variables.

— Dilip Sarwate

In my case, X and Y are jointly normal, I will try to explain why, please correct me if I am wrong. Suppose there is a set of independent univariate normal r.v.'s. Each element of X and Y is an arbitrary linear combination of these univariate variables from the set. Therefore, since the initial variables are independent and only linear transformations are involved, the resulting vectors X, Y, and Z are all multivariate normal r.v.'s. It follows the definition of a multivariate normal r.v., where

a^{T} X

$a^T X$ should be a univariate normal r.v. for any vector

a

$a$ . Does it make sense?

— Ivan

@Ivan Your explanation makes sense but the complaint is about the statement "Suppose we have two vectors of random variables, both are normal, i.e.,

X \sim N (μ_{X}, Σ_{X})

$X\sim N(\mu_X,\Sigma_X)$ and

Y \sim N (μ_{Y}, Σ_{Y})

$Y\sim N(\mu_Y,\Sigma_Y)$ " which does not mean that

X

$X$ and

Y

$Y$ are jointly normal. Nor does saying that "we know the correlation of any pair

(X_{i}, Y_{i})

$(X_i,Y_i)$ " mean that

X_{i}

$X_i$ and

Y_{i}

$Y_i$ are jointly normal even though, as you correctly state,

X \sim N (μ_{X}, Σ_{X})

$X\sim N(\mu_X,\Sigma_X)$ implies that

X_{i}

$X_i$ is normal (and similarly for

Y_{i}

$Y_i$ .) Univariate normality does not imply joint normality. See reference below.

— Dilip Sarwate

@Ivan See the discussion following this question

— Dilip Sarwate