Каково распределение суммы неидеальных гауссовых переменных?

Если $X$ распределен $N(\mu_X, \sigma^2_X)$ , $Y$ распределен $N(\mu_Y, \sigma^2_Y)$ и $Z = X + Y$ , я знаю , что $Z$ распределен $N(\mu_X + \mu_Y, \sigma^2_X + \sigma^2_Y)$ если X и Y независимы.

Но что произойдет, если X и Y не будут независимыми, то есть $(X, Y) \approx N\big( (\begin{smallmatrix} \mu_X\\\mu_Y \end{smallmatrix}) , (\begin{smallmatrix} \sigma^2_X && \sigma_{X,Y}\\ \sigma_{X,Y} && \sigma^2_Y \end{smallmatrix}) \big)$

Повлияет ли это на распределение суммы $Z$ ?

normal-distribution mathematical-statistics JCWong
источник

Просто хотелось бы отметить, что все виды совместных распределений

другой , чем двумерным нормальным , что до сих пор

незначительно нормально. И это различие будет иметь огромное значение для ответов.

(X, Y)

$(X,Y)$

X

$X$

Y

$Y$

@ G.JayKerns Я согласен с тем, что если

нормальные, но не обязательно совместно нормальные, то

может иметь распределение, отличное от нормального. Но заявление о том , что OP "

распределен

, если

независимы." абсолютно правильно. Если

X

$X$

Y

$Y$

X + Y

$X+Y$

Z

$Z$

N (μ_{x} + μ_{y}, σ_{x}^{2} + σ_{y}^{2})

$N(\mu_x + \mu_y, \sigma^2_x + \sigma^2_y)$

X

$X$

Y

$Y$

X

$X$

Y

$Y$ являются незначительно нормальными (как говорится в первой части предложения) и независимыми (согласно предположению во второй части предложения), тогда они также в целом нормальны. В вопросе OP совместная нормальность предполагается явно, и поэтому любая линейная комбинация

является нормальной.

X

$X$

Y

$Y$

Дилип Сарват

@Dilip, let me be clear that there is nothing wrong with the question and there is nothing wrong with your answer (+1) (or probability's, either (+1)). I was simply pointing out that if

X

$X$ and

Y

$Y$ are dependent then it isn't necessary that they are jointly normal, and it wasn't clear that the OP had considered that possibility. In addition, I am afraid (though I haven't spent a lot of time thinking) that without some other assumptions (like joint normality) the question might even be unanswerable.

As @G.JayKerns mentions, of course we can get all sorts of interesting behavior if we consider marginally, but not jointly, distributed normals. Here is a simple example: Let

X

$X$ be standard normal and

ε = \pm 1

$\varepsilon = \pm 1$ with probability 1/2 each, independently of

X

$X$ . Let

Y = ε X

$Y = \varepsilon X$ . Then

Y

$Y$ is also standard normal, but

Z = X + Y

$Z = X+Y$ is exactly equal to zero with probability 1/2 and is equal to

2 X

$2X$ with probability 1/2.

cardinal

We can get a whole variety of different behaviors by considering the bivariate copula that is associated with

(X, Y)

$(X,Y)$ via Sklar's theorem. If we use the Gaussian copula, then we get

(X, Y)

$(X,Y)$ are jointly normal, and so

Z = X + Y

$Z = X + Y$ is normally distributed. If the copula is not the Gaussian copula, then

X

$X$ and

Y

$Y$ are each still marginally distributed as normals, but are not jointly normal and so the sum will not be normally distributed, in general.

cardinal

Ответы:

See my comment on probabilityislogic's answer to this question. Here,

\begin{aligned} X + Y & \sim N (μ_{X} + μ_{Y}, σ_{X}^{2} + σ_{Y}^{2} + 2 σ_{X, Y}) \\ a X + b Y & \sim N (a μ_{X} + b μ_{Y}, a^{2} σ_{X}^{2} + b^{2} σ_{Y}^{2} + 2 a b σ_{X, Y}) \end{aligned}

$\begin{align*} X + Y &\sim N(\mu_X + \mu_Y,\; \sigma_X^2 + \sigma_Y^2 + 2\sigma_{X,Y})\\ aX + bY &\sim N(a\mu_X + b\mu_Y,\; a^2\sigma_X^2 + b^2\sigma_Y^2 + 2ab\sigma_{X,Y}) \end{align*}$ where

σ_{X, Y}

$\sigma_{X,Y}$ is the covariance of

X

$X$ and

Y

$Y$ . Nobody writes the off-diagonal entries in the covariance matrix as

σ_{x y}^{2}

$\sigma_{xy}^2$ as you have done. The off-diagonal entries are covariances which can be negative.

Dilip Sarwate
источник

@Kodiologist Thanks! I am surprised that the typos remained unnoticed for more than 4 years.

Dilip Sarwate

@dilip's answer is sufficient, but I just thought I'd add some details on how you get to the result. We can use the method of characteristic functions. For any $d$ -dimensional multivariate normal distribution $X\sim N_{d}(\mu,\Sigma)$ where $\mu=(\mu_1,\dots,\mu_d)^T$ and $\Sigma_{jk}=cov(X_j,X_k)\;\;j,k=1,\dots,d$ , the characteristic function is given by:

φ_{X} (t) = E [\exp (i t^{T} X)] = \exp (i t^{T} μ - \frac{1}{2} t^{T} Σ t)

$\varphi_{X}({\bf{t}})=E\left[\exp(i{\bf{t}}^TX)\right]=\exp\left(i{\bf{t}}^T\mu-\frac{1}{2}{\bf{t}}^T\Sigma{\bf{t}}\right)$

= \exp (i \sum_{j = 1}^{d} t_{j} μ_{j} - \frac{1}{2} \sum_{j = 1}^{d} \sum_{k = 1}^{d} t_{j} t_{k} Σ_{j k})

$=\exp\left(i\sum_{j=1}^{d}t_j\mu_j-\frac{1}{2}\sum_{j=1}^{d}\sum_{k=1}^{d}t_jt_k\Sigma_{jk}\right)$

For a one-dimensional normal variable $Y\sim N_1(\mu_Y,\sigma_Y^2)$ we get:

φ_{Y} (t) = \exp (i t μ_{Y} - \frac{1}{2} t^{2} σ_{Y}^{2})

$\varphi_Y(t)=\exp\left(it\mu_Y-\frac{1}{2}t^2\sigma_Y^2\right)$

Now, suppose we define a new random variable $Z={\bf{a}}^TX=\sum_{j=1}^{d}a_jX_j$ . For your case, we have $d=2$ and $a_1=a_2=1$ . The characteristic function for $Z$ is the basically the same as that for $X$ .

φ_{Z} (t) = E [\exp (i t Z)] = E [\exp (i t a^{T} X)] = φ_{X} (t a)

$\varphi_{Z}(t)=E\left[\exp(itZ)\right]=E\left[\exp(it{\bf{a}}^TX)\right]=\varphi_{X}(t{\bf{a}})$

= \exp (i t \sum_{j = 1}^{d} a_{j} μ_{j} - \frac{1}{2} t^{2} \sum_{j = 1}^{d} \sum_{k = 1}^{d} a_{j} a_{k} Σ_{j k})

$=\exp\left(it\sum_{j=1}^{d}a_j\mu_j-\frac{1}{2}t^2\sum_{j=1}^{d}\sum_{k=1}^{d}a_ja_k\Sigma_{jk}\right)$

If we compare this characteristic function with the characteristic function $\varphi_Y(t)$ we see that they are the same, but with $\mu_Y$ being replaced by $\mu_Z=\sum_{j=1}^{d}a_j\mu_j$ and with $\sigma_Y^2$ being replaced by $\sigma^2_Z=\sum_{j=1}^{d}\sum_{k=1}^{d}a_ja_k\Sigma_{jk}$ . Hence because the characteristic function of $Z$ is equivalent to the characteristic function of $Y$ , the distributions must also be equal. Hence $Z$ is normally distributed. We can simplify the expression for the variance by noting that $\Sigma_{jk}=\Sigma_{kj}$ and we get:

σ_{Z}^{2} = \sum_{j = 1}^{d} a_{j}^{2} Σ_{j j} + 2 \sum_{j = 2}^{d} \sum_{k = 1}^{j - 1} a_{j} a_{k} Σ_{j k}

$\sigma^2_Z=\sum_{j=1}^{d}a_j^2\Sigma_{jj}+2\sum_{j=2}^{d}\sum_{k=1}^{j-1}a_ja_k\Sigma_{jk}$

This is also the general formula for the variance of a linear combination of any set of random variables, independent or not, normal or not, where $\Sigma_{jj}=var(X_j)$ and $\Sigma_{jk}=cov(X_j,X_k)$ . Now if we specialise to $d=2$ and $a_1=a_2=1$ , the above formula becomes:

σ_{Z}^{2} = \sum_{j = 1}^{2} (1)^{2} Σ_{j j} + 2 \sum_{j = 2}^{2} \sum_{k = 1}^{j - 1} (1) (1) Σ_{j k} = Σ_{11} + Σ_{22} + 2 Σ_{21}

$\sigma^2_Z=\sum_{j=1}^{2}(1)^2\Sigma_{jj}+2\sum_{j=2}^{2}\sum_{k=1}^{j-1}(1)(1)\Sigma_{jk}=\Sigma_{11}+\Sigma_{22}+2\Sigma_{21}$

probabilityislogic
источник

+1 Thanks for taking the time to write out the details. Can this question be made part of the FAQ?

Dilip Sarwate