Hi @Payne001,
Thanks for posting here. It is indeed possible to have zero first-order Sobol’ indices but non-zero higher-order and total Sobol’ indices.
Let us define a deterministic computational model \mathcal{M}:\mathbb{R}^M\rightarrow \mathbb{R}, which maps the input vector \boldsymbol{x}\in \mathbb{R}^M to the model response y = \mathcal{M}(\boldsymbol{x}). In the context of global sensitivity analysis, each input component is modeled by a random variable. Hence, \boldsymbol{X} is a random vector, and the output is a random variable Y. Finally, we denote \boldsymbol{X}_{\boldsymbol{\mathsf{w}}} the subvector of \boldsymbol{X} defined by \boldsymbol{\mathsf{w}}\subset\{1,\ldots,M\}.
Recall the Hoeffding–Sobol decomposition of \mathcal{M}:
\begin{aligned}
\mathcal{M}(\boldsymbol{x}) &= \mathcal{M}_0 + \sum_{i}\mathcal{M}_i(x_i)+\sum_{i<j}\mathcal{M}_{i,j}(x_{i},x_{j})+\ldots+\mathcal{M}_{1,\ldots,M}(x_1,\ldots,x_M) \\
&=\sum_{\boldsymbol{\mathsf{w}} \subset \{1,\ldots,M\}}\mathcal{M}_{\boldsymbol{\mathsf{w}}}(\boldsymbol{x}_{\boldsymbol{\mathsf{w}}}),
\end{aligned}
where each component is given by
\mathcal{M}_{\boldsymbol{\mathsf{w}}}(\boldsymbol{x}_{\boldsymbol{\mathsf{w}}}) = \sum_{\boldsymbol{\mathsf{t}} \subset \boldsymbol{\mathsf{w}}} (-1)^{|\boldsymbol{\mathsf{w}}| - |{\boldsymbol{\mathsf{t}}}| } \mathbb{E}\left[Y \mid \boldsymbol{X}_{\boldsymbol{\mathsf{t}}} = \boldsymbol{x}_{\boldsymbol{\mathsf{t}}} \right].
This decomposition allows for decomposing the variance of Y to
\mathrm{Var}[Y] = \mathrm{Var}\left[\mathcal{M}(\boldsymbol{X})\right] = \sum_{\boldsymbol{\mathsf{w}} \subset \{1,\ldots,M\}}\mathrm{Var}\left[\mathcal{M}_{\boldsymbol{\mathsf{w}}}(\boldsymbol{X}_{\boldsymbol{\mathsf{w}}})\right].
The Sobol’ index is defined by
S_{\boldsymbol{\mathsf{w}}} = \frac{\mathrm{Var}\left[\mathcal{M}_{\boldsymbol{\mathsf{w}}}(X_{\boldsymbol{\mathsf{w}}})\right]}{\mathrm{Var}\left[\mathcal{M}(\boldsymbol{X})\right]},
For {\boldsymbol{\mathsf{w}}} containing a single element, say {\boldsymbol{\mathsf{w}}}=\{i\}, we obtain the first-order Sobol’ index for X_i:
S_i = \frac{\mathrm{Var}\left[\mathcal{M}_i(X_i)\right]}{\mathrm{Var}\left[\mathcal{M}(\boldsymbol{X})\right]}.
Similarly, we define the second-order Sobol’ index associated with X_i and X_j to quantify their interactive effect by taking {\boldsymbol{\mathsf{w}}}=\{i,j\}, that is,
S_{i,j} = \frac{\mathrm{Var}\left[\mathcal{M}_{i,j}(X_i,X_j)\right]}{\mathrm{Var}\left[\mathcal{M}(\boldsymbol{X})\right]}.
The total Sobol’ index for X_i aggregates all the effects related to X_i and is defined by
S_{T_i} = \sum_{i \in \boldsymbol{\mathsf{w}}\subset\{1,\ldots,M\}} S_{\boldsymbol{\mathsf{w}}}.
With the reminder above, let us consider a computational model defined by
\mathcal{M}(\boldsymbol{x}) = c_1x_1 + c_2x_2 + c_{3}x_1x_2,
and the input variables are independent standard Gaussian random variables, i.e., X_1,X_2\sim\mathcal{N}(0,1). Under this setup, the decomposition is given by
\mathcal{M}_1(x_1) = c_1x_1,\; \mathcal{M}_2(x_2) = c_2x_2,\; \mathcal{M}_{1,2}(x_1,x_2) = c_3x_1x_2.\;
Therefore, the first- and second-order Sobol’ indices are
S_1 = \frac{c^2_1}{c^2_1+c^2_2+c^2_3},\; S_2 = \frac{c^2_2}{c^2_1+c^2_2+c^2_3},\; S_{1,2} = \frac{c^2_3}{c^2_1+c^2_2+c^2_3}.
In addition, the total indices are
S_{T_1} = \frac{c^2_1 + c^2_3}{c^2_1+c^2_2+c^2_3},\; S_{T_2} = \frac{c^2_2 + c^2_3}{c^2_1+c^2_2+c^2_3}.
As a result, by taking c_1 = 0 and c_3\neq0, the first-order Sobol’ index for X_1 is zero whereas the second-order and total indices are non-zero, which concludes our analysis.
I hope this helps.