Structural reliability analysis based on analytical maximum entropy method using polynomial chaos expansion

Abstract

The maximum entropy (ME) method is a promising tool for structural reliability analysis by estimating the unknown probability density function (PDF) of given model response from its moment constraints. However, the classic ME algorithm has to resort to an iterative procedure due to non-linear constraints, and the required high order moment estimations may have large statistical error. In this paper, we (i) propose an analytical ME method based on integration by parts algorithm to transform the non-linear constraints to a system of linear equations and (ii) derive the polynomial chaos expansion (PCE) multiplication for improving higher order moment calculation required in the previous step efficiently. Thus, an analytical formula of response PDF is obtained directly without intensively iterative procedure and associated convergence error, and it is followed by probability failure estimation using numerical integration computation. Two structural engineering cases are implemented to illustrate the accuracy and efficiency of the proposed method.

Appendices

Appendix 1

1.1 Chebyshev polynomials

The definition of Chebyshev polynomials is

$$ {\displaystyle \begin{array}{l}{T}_0(x)=1,\\ {}{T}_n(x)=\frac{n}{2}\sum \limits_{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor }{\left(-1\right)}^k\frac{\left(n-k-1\right)!}{k!\left(n-2k\right)!}{(2x)}^{n-2k},\kern0.6em n\ge 1\end{array}} $$

(38)

and the most well-known form of Chebyshev polynomials is given by the recurrence relation

$$ {\displaystyle \begin{array}{l}{T}_0(x)=1,\\ {}{T}_1(x)=x,\\ {}{T}_{n+1}(x)=2{xT}_n(x)-{T}_{n-1}(x),\kern0.8000001em n\ge 2\end{array}} $$

(39)

where x∈[− 1, 1]. This definition is also referred as Chebyshev polynomials of the first kind, and there is another definition of Chebyshev polynomials of the second kind:

$$ {\displaystyle \begin{array}{l}{U}_0(x)=1,\\ {}{U}_1(x)=2x,\\ {}{U}_{n+1}(x)=2{xU}_n(x)-{U}_{n-1}(x),\kern0.8000001em n\ge 2\end{array}} $$

(40)

The relationship between these two kinds of Chebyshev polynomials are given as (41)-(44):

$$ \frac{dT_n(x)}{dx}={nU}_{n-1}(x) $$

(41)

$$ {T}_j(x){U}_k(x)=\left\{\begin{array}{cc}\frac{1}{2}\left({U}_{j+k}(x)+{U}_{k-j}(x)\right),& k\ge j-1,\\ {}\frac{1}{2}\left({U}_{j+k}(x)-{U}_{j-k-2}(x)\right),& k\le j-2.\end{array}\right. $$

(42)

with the convention U₋₁ ≡ 0.

$$ {T}_n(x)=\frac{1}{2\left(n+1\right)}\frac{dT_{n+1}(x)}{dx}-\frac{1}{2\left(n-1\right)}\frac{dT_{n-1}(x)}{dx},\kern0.9000001em \forall i\ge 1. $$

(43)

$$ \left(1-{x}^2\right){U}_{n-1}(x)={xT}_n(x)-{T}_{n+1}(x)=\frac{T_{n-1}(x)-{T}_{n+1}(x)}{2} $$

(44)

1.2 Hermite polynomials

The Hermite polynomials {H_n(x)}_n ≥ 0 are defined as

$$ {H}_n(x)={\left(-1\right)}^n{e}^{\frac{x^2}{2}}\frac{d^n}{dx^n}\left({e}^{-\frac{x^2}{2}}\right) $$

(45)

and the well-known recursive relation of the Hermite polynomials is

$$ \left\{\begin{array}{l}{H}_0(x)\equiv 1,\\ {}{H}_1(x)=2x,\\ {}{H}_{n+1}(x)=2{xH}_n(x)-2{nH}_{n-1}(x)\kern0.2em \end{array}\right. $$

(46)

Appendix 2

In this section, the derivation details of (12) is provided.

Consider (10), and take ${F}_1^{(i)}(y)={\int}_{-1}^1\left(1-{y}^2\right)\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right)\frac{1}{2\left(i+1\right)}{dT}_{i+1}(y)$ and ${F}_2^{(i)}(y)={\int}_{-1}^1\left(1-{y}^2\right)\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right)\frac{1}{2\left(i-1\right)}{dT}_{i-1}(y)$ for simplicity. Then, we would solve the two parts separately.

First of all, F₁⁽ⁱ⁾(x) could be transformed as

$$ {\displaystyle \begin{array}{l}{F}_1^{(i)}(y)={\left[\frac{1}{2\left(i+1\right)}\left(1-{y}^2\right)\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right)\right]}_{-1}^1\\ {}\kern3.099999em -{\int}_{-1}^1\frac{1}{2\left(i+1\right)}{T}_{i+1}(y)\left[-2x+\left(1-{y}^2\right)\frac{d}{dy}\left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right)\right]\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right) dy\\ {}\kern2.4em =\frac{1}{i+1}{\int}_{-1}^1{yT}_{i+1}(y)\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right) dy\\ {}\kern3.099999em +\frac{1}{2\left(i+1\right)}{\int}_{-1}^1{T}_{i+1}(y)\left(1-{y}^2\right)\frac{d}{dy}\left(\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right)\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right) dy\end{array}} $$

(47)

Take (39) into consideration, the first integral term above is derived as

$$ {\displaystyle \begin{array}{l}\frac{1}{i+1}{\int}_{-1}^1{yT}_{i+1}(y)\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right) dy\\ {}=\frac{1}{i+1}{\int}_{-1}^1\frac{T_{i+1}(y)+{T}_i(y)}{2}\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right) dy\\ {}=\frac{1}{2\left(i+1\right)}\left({M}_{i+2}+{M}_i\right)\end{array}} $$

(48)

Then, take (41) and (42) into consideration, the second integral term in (47) is derived as

$$ {\displaystyle \begin{array}{l}\frac{1}{2\left(i+1\right)}{\int}_{-1}^1{T}_{i+1}(y)\left(1-{y}^2\right)\frac{d}{dy}\left(\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right)\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right) dy\\ {}=\frac{1}{2\left(i+1\right)}{\int}_{-1}^1{T}_{i+1}(y)\left(1-{y}^2\right)\left(\sum \limits_{j=1}^mj{\lambda}_j{U}_{j-1}(y)\right)\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(x)\right) dy\\ {}=\frac{1}{2\left(i+1\right)}{\int}_{-1}^1\left(1-{y}^2\right)\left(\sum \limits_{j=1}^mj{\lambda}_j{T}_{i+1}(y){U}_{j-1}(y)\right)\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right) dy\\ {}=\frac{1}{2\left(i+1\right)}{\int}_{-1}^1\left(1-{y}^2\right)\left[\begin{array}{l}\left(\sum \limits_{j=1}^ij{\lambda}_j\frac{U_{i+j}(y)-{U}_{i-j}(y)}{2}\right)+\left(i+1\right){\lambda}_{i+1}\frac{U_{2i+1}}{2}\\ {}+\left(\sum \limits_{j=i+2}^mj{\lambda}_j\frac{U_{i+j}(y)+{U}_{j-i-2}(y)}{2}\right)\end{array}\right]\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right) dy\end{array}} $$

(49)

where the Chebyshev polynomials of the second kind come as intermediate polynomials.

By using (44), the polynomial multiplication of Chebyshev polynomials in (49) is given as the following:

$$ {\displaystyle \begin{array}{l}\left(1-{y}^2\right)\frac{U_{i+j}(y)-{U}_{i-j}(y)}{2}=\frac{T_{i+j}(y)-{T}_{i+j+2}(y)}{2}-\frac{T_{i-j}(y)-{T}_{i-j+2}(y)}{2}\\ {}\left(1-{y}^2\right)\frac{U_{2i+1}}{2}=\frac{T_{2i+1}(y)-{T}_{2i+3}(y)}{2}\\ {}\left(1-{y}^2\right)\frac{U_{i+j}(y)+{U}_{j-i-2}(y)}{2}=\frac{T_{i+j}(y)-{T}_{i+j+2}(y)}{2}+\frac{T_{j-i-2}(y)-{T}_{j-i}(y)}{2}\end{array}} $$

(50)

Substitute (50) into (49), the second integral term in (47) is derived as follows:

$$ {\displaystyle \begin{array}{l}\frac{1}{2\left(i+1\right)}{\int}_{-1}^1{T}_{i+1}(y)\left(1-{y}^2\right)\frac{d}{dy}\left(\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right)\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right) dy\\ {}=\frac{1}{2\left(i+1\right)}{\int}_{-1}^1\left[\sum \limits_{j=1}^i\frac{j}{2}{\lambda}_j\left(\frac{T_{i+j}(y)-{T}_{i+j+2}(y)}{2}-\frac{T_{i-j}(y)-{T}_{i-j+2}(y)}{2}\right)\right]\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right) dy\\ {}\kern0.7em +\frac{1}{2\left(i+1\right)}{\int}_{-1}^1\frac{i+1}{2}{\lambda}_{i+1}\frac{T_{2i+1}(y)-{T}_{2i+3}(y)}{2}\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right) dy\\ {}\kern0.7em +\frac{1}{2\left(i+1\right)}{\int}_{-1}^1\sum \limits_{j=i+2}^m\frac{j}{2}{\lambda}_j\left(\frac{T_{i+j}(y)-{T}_{i+j+2}(y)}{2}+\frac{T_{j-i-2}(y)-{T}_{j-i}(y)}{2}\right)\exp \left(-1-\sum \limits_{j=0}^m{\lambda}_j{T}_j(y)\right) dy\\ {}=\frac{1}{8\left(i+1\right)}\sum \limits_{j=1}^ij\left({M}_{i+j}-{M}_{i+j+2}-{M}_{i-j}+{M}_{i-j+2}\right){\lambda}_j+\frac{1}{8}\left({M}_{2i+1}-{M}_{2i+3}\right){\lambda}_{i+1}\\ {}\kern0.9000001em +\frac{1}{8\left(i+1\right)}\sum \limits_{j=i+2}^mj\left({M}_{i+j}-{M}_{i+j+2}+{M}_{j-i-2}-{M}_{j-i}\right){\lambda}_j\end{array}} $$

(51)

By Combining (48) and (51), we have the following relationship:

$$ {\displaystyle \begin{array}{l}{F}_1^{(i)}(y)=\frac{1}{2\left(i+1\right)}\left({M}_{i+2}+{M}_i\right)+\frac{1}{8\left(i+1\right)}\sum \limits_{j=1}^ij\left({M}_{i+j}-{M}_{i+j+2}-{M}_{i-j}+{M}_{i-j+2}\right){\lambda}_j\\ {}\kern3.099999em +\frac{1}{8}\left({M}_{2i+1}-{M}_{2i+3}\right){\lambda}_{i+1}+\frac{1}{8\left(i+1\right)}\sum \limits_{j=i+2}^mj\left({M}_{i+j}-{M}_{i+j+2}+{M}_{j-i-2}-{M}_{j-i}\right){\lambda}_j\end{array}} $$

(52)

Secondly, when it comes to F₂⁽ⁱ⁾(x), it is a little more complicated. When i = 1, since T₀(x) is a constant according to the definition of Chebyshev polynomials. Otherwise, when i ≥ 2, we could derive F₂⁽ⁱ⁾(x) with a similar procedure as (47)-(51) with the following result:

$$ {\displaystyle \begin{array}{l}{F}_2^{(i)}(y)=\frac{1}{2\left(i-1\right)}\left({M}_i+{M}_{i-2}\right)+\frac{1}{8\left(i-1\right)}\sum \limits_{j=1}^{i-2}j\left({M}_{i+j-2}-{M}_{i+j}-{M}_{i-j-2}+{M}_{i-j}\right){\lambda}_j\\ {}\kern3.099999em +\frac{1}{8}\left({M}_{2i-3}-{M}_{2i-1}\right){\lambda}_{i+1}+\frac{1}{8\left(i-1\right)}\sum \limits_{j=i+2}^mj\left({M}_{i+j-2}-{M}_{i+j}+{M}_{j-i}-{M}_{j-i+2}\right){\lambda}_j\end{array}} $$

(53)

What is more, consider (8) and (9), we would establish the following equations. When i = 1, we have

$$ {\displaystyle \begin{array}{l}{\int}_{-1}^1\left(1-{y}^2\right){T}_1(y)\exp \left(-1-\sum \limits_{j=1}^m{\lambda}_j{T}_j(y)\right) dy={F}_1^{(1)}(y)\\ {}{\int}_{-1}^1\left(1-{y}^2\right){T}_i(y)\exp \left(-1-\sum \limits_{j=1}^m{\lambda}_j{T}_j(y)\right) dy={F}_1^{(i)}(y)-{F}_2^{(i)}(y)\end{array}} $$

(54)

and it could be rewritten as

$$ \frac{1}{16}\left(2{M}_2-{M}_4-{M}_0\right){\lambda}_1+\frac{1}{8}\left({M}_3-{M}_5\right){\lambda}_2+\frac{1}{16}\sum \limits_{j=3}^mj\left({M}_{j+1}-{M}_{j+3}+{M}_{j-3}-{M}_{j-1}\right){\lambda}_j=\frac{1}{4}\left({M}_1-{M}_3\right) $$

(55)

and for i ≥ 2

$$ {\displaystyle \begin{array}{l}\frac{1}{8\left(i+1\right)}\sum \limits_{j=1}^ij\left({M}_{i+j}-{M}_{i+j+2}-{M}_{i-j}+{M}_{i-j+2}\right){\lambda}_j+\frac{1}{8}\left({M}_{2i+1}-{M}_{2i+3}\right){\lambda}_{i+1}\kern0.1em \\ {}+\frac{1}{8\left(i+1\right)}\sum \limits_{j=i+2}^mj\left({M}_{i+j}-{M}_{i+j+2}+{M}_{j-i-2}-{M}_{j-i}\right){\lambda}_j-\frac{1}{8\left(i-1\right)}\sum \limits_{j=1}^{i-2}j\left({M}_{i+j-2}-{M}_{i+j}-{M}_{i-j-2}+{M}_{i-j}\right){\lambda}_j\\ {}-\frac{1}{8}\left({M}_{2i-3}-{M}_{2i-1}\right){\lambda}_{i+1}-\frac{1}{8\left(i-1\right)}\sum \limits_{j=i+2}^mj\left({M}_{i+j-2}-{M}_{i+j}+{M}_{j-i}-{M}_{j-i+2}\right){\lambda}_j\\ {}=\left(\frac{1}{2}-\frac{1}{2\left(i+1\right)}+\frac{1}{2\left(i-1\right)}\right){M}_i-\left(\frac{1}{4}+\frac{1}{2\left(i+1\right)}\right){M}_{i+2}-\left(\frac{1}{4}-\frac{1}{2\left(i-1\right)}\right){M}_{i-2}\end{array}} $$

(56)

Appendix 3

In this section, we would prove the formula of two PCEs multiplication.

First of all, for any non-negative integer α and β, the product of two Hermite polynomials can be expressed as (Savin and Faverjon 2017)

$$ {H}_{\alpha }(x){H}_{\beta }(x)=\sum \limits_{r\le \min \left(\alpha, \beta \right)}{\left[\left(\begin{array}{l}\alpha \\ {}r\end{array}\right)\left(\begin{array}{l}\beta \\ {}r\end{array}\right)\left(\begin{array}{c}\alpha +\beta -2r\\ {}\alpha -r\end{array}\right)\right]}^{\frac{1}{2}}{H}_{\alpha +\beta -2r}(x) $$

(57)

Then, this formula could be generalized to multi-dimensional form as (58).

$$ {\psi}_{\boldsymbol{\alpha}}\left(\boldsymbol{\xi} \right){\psi}_{\boldsymbol{\beta}}\left(\boldsymbol{\xi} \right)=\sum \limits_{r\le \min \left(\boldsymbol{\alpha}, \boldsymbol{\beta} \right)}B\left(\boldsymbol{\alpha}, \boldsymbol{\beta}, \boldsymbol{r}\right){\psi}_{\boldsymbol{\alpha} +\boldsymbol{\beta} -2\boldsymbol{r}}\left(\boldsymbol{\xi} \right) $$

(58)

where $B\left(\boldsymbol{\alpha}, \boldsymbol{\beta}, \boldsymbol{r}\right)={\left[\left(\begin{array}{l}\boldsymbol{\alpha} \\ {}\boldsymbol{r}\end{array}\right)\left(\begin{array}{l}\boldsymbol{\beta} \\ {}\boldsymbol{r}\end{array}\right)\left(\begin{array}{c}\boldsymbol{\alpha} +\boldsymbol{\beta} -2\boldsymbol{r}\\ {}\boldsymbol{\alpha} -\boldsymbol{r}\end{array}\right)\right]}^{\frac{1}{2}}$, and the meaning of the notations above is shown in Section 3.2.

With the preparation, we could derive the multiplication of two PCEs as follows.

Suppose u and v have PCE formula with the same n-dimensional standardized random variables ξ = (ξ₁, ⋯, ξ_n)^T but different order p_α and p_β respectively. Namely, we have $u=\sum \limits_{\left|\boldsymbol{\alpha} \right|\le {p}_{\alpha }}{u}_{\boldsymbol{\alpha}}{\psi}_{\boldsymbol{\alpha}}\left(\boldsymbol{\xi} \right)$, $v=\sum \limits_{\left|\boldsymbol{\beta} \right|\le {p}_{\beta }}{v}_{\boldsymbol{\beta}}{\psi}_{\boldsymbol{\beta}}\left(\boldsymbol{\xi} \right)$. So the multiplication of u and v is

$$ {\displaystyle \begin{array}{l} uv=\sum \limits_{\left|\boldsymbol{\alpha} \right|\le {p}_{\alpha }}\sum \limits_{\left|\boldsymbol{\beta} \right|\le {p}_{\beta }}{u}_{\boldsymbol{\alpha}}{v}_{\boldsymbol{\beta}}{\psi}_{\boldsymbol{\alpha}}\left(\boldsymbol{\xi} \right){\psi}_{\boldsymbol{\beta}}\left(\boldsymbol{\xi} \right)\\ {}=\sum \limits_{\left|\boldsymbol{\alpha} \right|\le {p}_{\alpha }}\sum \limits_{\left|\boldsymbol{\beta} \right|\le {p}_{\beta }}{u}_{\boldsymbol{\alpha}}{v}_{\boldsymbol{\beta}}\sum \limits_{r\le \min \left(\boldsymbol{\alpha}, \boldsymbol{\beta} \right)}B\left(\boldsymbol{\alpha}, \boldsymbol{\beta}, \boldsymbol{r}\right){\psi}_{\boldsymbol{\alpha} +\boldsymbol{\beta} -2\boldsymbol{r}}\left(\boldsymbol{\xi} \right)\\ {}=\sum \limits_{\left|\boldsymbol{\alpha} \right|\le {p}_{\alpha }}\sum \limits_{\left|\boldsymbol{\beta} \right|\le {p}_{\beta }}{u}_{\boldsymbol{\alpha}}{v}_{\boldsymbol{\beta}}\sum \limits_{\boldsymbol{r}\le \min \left(\boldsymbol{\alpha}, \boldsymbol{\beta} \right)}\left(\begin{array}{c}\boldsymbol{\alpha} \\ {}\boldsymbol{r}\end{array}\right)\left(\begin{array}{c}\boldsymbol{\beta} \\ {}\boldsymbol{r}\end{array}\right)r!\frac{\sqrt{\left(\boldsymbol{\alpha} +\boldsymbol{\beta} -2\boldsymbol{r}\right)!}}{\sqrt{\boldsymbol{\alpha} !\boldsymbol{\beta} !}}{\psi}_{\boldsymbol{\alpha} +\boldsymbol{\beta} -2\boldsymbol{r}}\left(\boldsymbol{\xi} \right)\end{array}} $$

(59)

Let $\tilde{\boldsymbol{\alpha}}=\boldsymbol{\alpha} -\boldsymbol{r}$, $\tilde{\boldsymbol{\beta}}=\boldsymbol{\beta} -\boldsymbol{r}$, then r = min(α, β) is equivalent to $\tilde{\boldsymbol{\alpha}},\tilde{\boldsymbol{\beta}}\ge 0$. Alternatively, $\boldsymbol{\alpha} =\tilde{\boldsymbol{\alpha}}+\boldsymbol{r}$, $\boldsymbol{\beta} =\tilde{\boldsymbol{\beta}}+\boldsymbol{r}$ and the above summation can be rewritten as

$$ uv=\sum \limits_{\left|\tilde{\boldsymbol{\alpha}}\right|\le {p}_{\alpha }}\sum \limits_{\left|\tilde{\boldsymbol{\beta}}\right|\le {p}_{\beta }}\sum \limits_{\begin{array}{l}\left|\boldsymbol{r}+\tilde{\boldsymbol{\alpha}}\right|\le {p}_{\alpha },\\ {}\left|\boldsymbol{r}+\tilde{\boldsymbol{\beta}}\right|\le {p}_{\beta}\end{array}}{u}_{\tilde{\boldsymbol{\alpha}}+\boldsymbol{r}}{v}_{\tilde{\boldsymbol{\beta}}+\boldsymbol{r}}\left(\begin{array}{c}\tilde{\boldsymbol{\alpha}}+\boldsymbol{r}\\ {}\boldsymbol{r}\end{array}\right)\left(\begin{array}{c}\tilde{\boldsymbol{\beta}}+\boldsymbol{r}\\ {}\boldsymbol{r}\end{array}\right)r!\frac{\sqrt{\left(\tilde{\boldsymbol{\alpha}}+\tilde{\boldsymbol{\beta}}\right)!}}{\sqrt{\left(\tilde{\boldsymbol{\alpha}}+\boldsymbol{r}\right)!\left(\tilde{\boldsymbol{\beta}}+\boldsymbol{r}\right)!}}{\psi}_{\tilde{\boldsymbol{\alpha}}+\tilde{\boldsymbol{\beta}}}\left(\boldsymbol{\xi} \right) $$

(60)

For simplicity, we still denote $\boldsymbol{\alpha} =\tilde{\boldsymbol{\alpha}}$ and $\boldsymbol{\beta} =\tilde{\boldsymbol{\beta}}$. Then, let θ = α + β, thus α = θ − β ≥ 0 and 0 ≤ β ≤ θ. The above summation is equivalent to

$$ {\displaystyle \begin{array}{l} uv=\sum \limits_{\left|\boldsymbol{\theta} \right|\le {p}_{\alpha }+{p}_{\beta }}\sum \limits_{\boldsymbol{\alpha} +\boldsymbol{\beta} =\boldsymbol{\theta}}\sum \limits_{\begin{array}{l}\left|\boldsymbol{\alpha} +\boldsymbol{r}\right|\le {p}_{\alpha },\\ {}\left|\boldsymbol{\beta} +\boldsymbol{r}\right|\le {p}_{\beta}\end{array}}{u}_{\boldsymbol{\alpha} +\boldsymbol{r}}{v}_{\boldsymbol{\beta} +\boldsymbol{r}}\left(\begin{array}{c}\boldsymbol{\alpha} +\boldsymbol{r}\\ {}\boldsymbol{r}\end{array}\right)\left(\begin{array}{c}\boldsymbol{\beta} +\boldsymbol{r}\\ {}\boldsymbol{r}\end{array}\right)r!\frac{\sqrt{\boldsymbol{\theta} !}}{\sqrt{\left(\boldsymbol{\alpha} +\boldsymbol{r}\right)!\left(\boldsymbol{\beta} +\boldsymbol{r}\right)!}}{\psi}_{\boldsymbol{\theta}}\left(\boldsymbol{\xi} \right)\\ {}=\sum \limits_{\left|\boldsymbol{\theta} \right|\le {p}_{\alpha }+{p}_{\beta }}\sum \limits_{0\le \boldsymbol{\beta} \le \boldsymbol{\theta}}\sum \limits_{\begin{array}{l}\left|\boldsymbol{\theta} -\boldsymbol{\beta} +\boldsymbol{r}\right|\le {p}_{\alpha },\\ {}\left|\boldsymbol{\beta} +\boldsymbol{r}\right|\le {p}_{\beta}\end{array}}{u}_{\boldsymbol{\theta} -\boldsymbol{\beta} +\boldsymbol{r}}{v}_{\boldsymbol{\beta} +\boldsymbol{r}}\left(\begin{array}{c}\boldsymbol{\theta} -\boldsymbol{\beta} +\boldsymbol{r}\\ {}\boldsymbol{r}\end{array}\right)\left(\begin{array}{c}\boldsymbol{\beta} +\boldsymbol{r}\\ {}\boldsymbol{r}\end{array}\right)r!\frac{\sqrt{\boldsymbol{\theta} !}}{\sqrt{\left(\boldsymbol{\theta} -\boldsymbol{\beta} +\boldsymbol{r}\right)!\left(\boldsymbol{\beta} +\boldsymbol{r}\right)!}}{\psi}_{\boldsymbol{\theta}}\left(\boldsymbol{\xi} \right)\end{array}} $$

(61)

For simplicity, denote

$$ C\left(\boldsymbol{\theta}, \boldsymbol{\beta}, \boldsymbol{r}\right)=\left(\begin{array}{c}\boldsymbol{\theta} -\boldsymbol{\beta} +\boldsymbol{r}\\ {}\boldsymbol{r}\end{array}\right)\left(\begin{array}{c}\boldsymbol{\beta} +\boldsymbol{r}\\ {}\boldsymbol{r}\end{array}\right)r!\frac{\sqrt{\boldsymbol{\theta} !}}{\sqrt{\left(\boldsymbol{\theta} -\boldsymbol{\beta} +\boldsymbol{r}\right)!\left(\boldsymbol{\beta} +\boldsymbol{r}\right)!}} $$

(62)

Thus, we have

$$ uv=\sum \limits_{\left|\boldsymbol{\theta} \right|\le {p}_{\alpha }+{p}_{\beta }}\sum \limits_{0\le \boldsymbol{\beta} \le \boldsymbol{\theta}}\sum \limits_{\begin{array}{l}\left|\boldsymbol{\theta} -\boldsymbol{\beta} +\boldsymbol{r}\right|\le {p}_{\alpha },\\ {}\left|\boldsymbol{\beta} +\boldsymbol{r}\right|\le {p}_{\beta}\end{array}}C\left(\boldsymbol{\theta}, \boldsymbol{\beta}, \boldsymbol{r}\right){u}_{\boldsymbol{\theta} -\boldsymbol{\beta} +\boldsymbol{r}}{v}_{\boldsymbol{\beta} +\boldsymbol{r}}{\psi}_{\boldsymbol{\theta}}\left(\boldsymbol{\xi} \right) $$

(63)

which completes the proof.

Appendix 4

In this section, we would prove the statement in Section 3.2 that the multiplication of two PCEs converges in the L₂ sense. Let us consider a system as y = g(x) with independent distribution input x. And assume that the PCE approximation of this system is expressed as y_p = g_PCE(ξ), where p is the order of PCE, and ξ is the standard normal variable associated with input x. Then, since PCE is L₂ convergence, we have

$$ \underset{p\to \infty }{\lim }E{\left({y}_p-y\right)}^2=\underset{p\to \infty }{\lim }{\int}_{\Omega}{\left({y}_p-y\right)}^2f\left(\boldsymbol{x}\right)d\boldsymbol{x}=0 $$

(64)

where Ω is the domain of input x, and f(x) is the joint PDF that subjects to ∫_Ωf(x)dx = 1.

Therefore, ∀ 0 < ε < 1, ∃ n > N where N is a positive integer,

$$ \left|{\int}_{\Omega}{\left({y}_p-y\right)}^2f\left(\boldsymbol{x}\right)d\boldsymbol{x}\right|<\varepsilon <1 $$

(65)

Now considering the fact that (y_p − y)² ≥ 0, if (y_p − y)² ≥ 1, we would derive that

$$ \left|{\int}_{\Omega}{\left({y}_p-y\right)}^2f\left(\boldsymbol{x}\right)d\boldsymbol{x}\right|\ge \left|{\int}_{\Omega}f\left(\boldsymbol{x}\right)d\boldsymbol{x}\right|=1 $$

(66)

which is in contradiction with (65). So, we would have 0 ≤ (y_p − y)² < 1. This means that |y_p − y| is bounded.

Noting that the absolute value of response of given system is less than 1 in our proposed method, namely, |y| ≤ 1. So, we could obtain that

$$ \left|{y}_p\right|\le 2 $$

(67)

Now, we would prove the convergence of PCE multiplication. First of all, we would prove a basis inequality as

$$ {\displaystyle \begin{array}{l}{\left({y}_p^k-{y}^k\right)}^2={\left({y}_p-y\right)}^2{\left({y}_p^{k-1}+{y}_p^{k-2}y+\cdots {y}_p{y}^{k-2}+{y}^{k-1}\right)}^2\\ {}\kern4.599998em \le {\left({y}_p-y\right)}^2{\left(\left|{y}_p^{k-1}\right|+\left|{y}_p^{k-2}\right|\left|y\right|+\cdots \left|{y}_p\right|\left|{y}^{k-2}\right|+\left|{y}^{k-1}\right|\right)}^2\\ {}\kern4.499998em \le {\left({y}_p-y\right)}^2{\left({2}^{k-1}+{2}^{k-2}+\cdots 2+1\right)}^2\\ {}\kern4.499998em ={\left({y}_p-y\right)}^2{\left({2}^k-1\right)}^2\end{array}} $$

(68)

where k means the order of required moment.

Then, with this preparation, we could derive that for a limited number k,

$$ {\displaystyle \begin{array}{l}\underset{p\to \infty }{\lim }E{\left({y}_p^k-{y}^k\right)}^2=\underset{p\to \infty }{\lim }{\int}_{\Omega}{\left({y}_p^k-{y}^k\right)}^2f\left(\boldsymbol{x}\right)d\boldsymbol{x}\\ {}\le \underset{p\to \infty }{\lim }{\int}_{\Omega}{\left({y}_p-y\right)}^2f\left(\boldsymbol{x}\right)d\boldsymbol{x}\\ {}\le {\left({2}^k-1\right)}^2\underset{p\to \infty }{\lim }{\int}_{\Omega}{\left({y}_p-y\right)}^2f\left(\boldsymbol{x}\right)d\boldsymbol{x}\\ {}=0\end{array}} $$

(69)

This means that if y_p is L₂ convergence with given system, the k-power of y_p also converges to y^k in L₂ sense. Thus, the approach of PCE multiplication could provide an accurate estimation of high order moment information. So, the proof completes.