Exposing Region Splicing Forgeries with Blind Local Noise Estimation

Abstract

Region splicing is a simple and common digital image tampering operation, where a chosen region from one image is composited into another image with the aim to modify the original image’s content. In this paper, we describe an effective method to expose region splicing by revealing inconsistencies in local noise levels, based on the fact that images of different origins may have different noise characteristics introduced by the sensors or post-processing steps. The basis of our region splicing detection method is a new blind noise estimation algorithm, which exploits a particular regular property of the kurtosis of nature images in band-pass domains and the relationship between noise characteristics and kurtosis. The estimation of noise statistics is formulated as an optimization problem with closed-form solution, and is further extended to an efficient estimation method of local noise statistics. We demonstrate the efficacy of our blind global and local noise estimation methods on natural images, and evaluate the performances and robustness of the region splicing detection method on forged images.

Appendices

Appendix 1: Lyapunov Central Limit Theorem

Suppose $\{X_1, \cdots , X_n\}$ is a sequence of independent random variables, each with finite expected value $\mu _i$ and variance $\sigma ^2_i$. Define $s_n^2 = \sum _{i=1}^n \sigma _i^2$, If for some $\delta > 0$, the ”Lyapunovs condition”

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{s_{n}^{2+\delta }} \sum _{i=1}^{n} \mathrm{E }\big [\,|X_{i} - \mu _{i}|^{2+\delta }\,\big ] = 0 \end{aligned}$$

is satisfied, then a sum of $(X_i - \mu _i)/s_n$ converges in distribution to a standard normal random variable, as $n$ goes to infinity:

$$\begin{aligned} \frac{1}{s_n} \sum _{i=1}^{n} (X_i - \mu _i) \ \xrightarrow {d}\ \mathcal {N}(0,\;1). \end{aligned}$$

Appendix 2: Derivation of Claim 1

Based on the joint density function of the GSM variable $\mathbf {x}$, Eq. (5), we can obtain the marginal distribution of its projection on a non-zero vector $\mathbf {w}$ as:

$$\begin{aligned}&\!\!\!\! p_\mathbf {w}(t) = \int _{\mathbf {x}:\mathbf {w}^{\scriptscriptstyle T}\mathbf {x}= t} p(\mathbf {x}) d\mathbf {x}= \\&\!\!\!\!\int _{\mathrm {z}} p_\mathrm {z}(\mathrm {z})d\mathrm {z}\int _{\mathbf {x}:\mathbf {w}^{\scriptscriptstyle T}\mathbf {x}= t} \!\! \frac{1}{\sqrt{(2\pi \mathrm {z})^d |\mathrm {det}(\Sigma _\mathbf {x})|}} \exp \left(\! -\frac{\mathbf {x}^{\scriptscriptstyle T}\Sigma _\mathbf {x}^{-1} \mathbf {x}}{2\mathrm {z}} \right) d\mathbf {x}. \end{aligned}$$

The marginalization in the inside integral reduces to a Gaussian distribution with zero mean and variance $\mathrm {z}\mathbf {w}^{\scriptscriptstyle T}\Sigma _\mathbf {x}\mathbf {w}$, based on the property of Gaussian distributions. Therefore, $p_\mathbf {w}(t)$ is a 1D GSM model with mixing density $p_\mathrm {z}(\mathrm {z})$, as:

$$\begin{aligned}&\!\!\! p_\mathbf {w}(t) = \int _{\mathrm {z}} \mathcal{N}_{t}(0,\mathrm {z}\mathbf {w}^{\scriptscriptstyle T}\Sigma _\mathbf {x}\mathbf {w}) p_\mathrm {z}(\mathrm {z})d\mathrm {z}= \\&\!\!\! \int _{\mathrm {z}} \frac{1}{\sqrt{2\pi \mathrm {z}\mathbf {w}^{\scriptscriptstyle T}\Sigma _\mathbf {x}\mathbf {w}}} \exp \left( -\frac{t^2}{2\mathrm {z}\mathbf {w}^{\scriptscriptstyle T}\Sigma _\mathbf {x}\mathbf {w}} \right) p_\mathrm {z}(\mathrm {z})d\mathrm {z}\end{aligned}$$

Now, the variance of $\mathbf {w}^{\scriptscriptstyle T}\mathbf {x}$ is computed as

$$\begin{aligned}&\!\!\! \mathcal{E}_{t}\left\{ t^2\right\} = \int _{\mathrm {z}} p_\mathrm {z}(\mathrm {z})d\mathrm {z}\int _t t^2\mathcal{N}_{t}(0,\mathrm {z}\mathbf {w}^{\scriptscriptstyle T}\Sigma _\mathbf {x}\mathbf {w}) = \\&\!\!\! \mathbf {w}^{\scriptscriptstyle T}\Sigma _\mathbf {x}\mathbf {w}\int _\mathrm {z}\mathrm {z}p_\mathrm {z}(\mathrm {z})d\mathrm {z}= \mathbf {w}^{\scriptscriptstyle T}\Sigma _\mathbf {x}\mathbf {w}\mathcal{E}_{\mathrm {z}}\left\{ \mathrm {z}\right\} . \end{aligned}$$

Furthermore, the fourth order moment of $\mathbf {w}^{\scriptscriptstyle T}\mathbf {x}$ is given as

$$\begin{aligned}&\!\!\! \mathcal{E}_{t}\left\{ t^4\right\} = \int _{\mathrm {z}} p_\mathrm {z}(\mathrm {z})d\mathrm {z}\int _t t^4\mathcal{N}_{t}(0,\mathrm {z}\mathbf {w}^{\scriptscriptstyle T}\Sigma _\mathbf {x}\mathbf {w}) = \\&\!\!\! 3 \mathbf {w}^{\scriptscriptstyle T}\Sigma _\mathbf {x}\mathbf {w}\int _\mathrm {z}\mathrm {z}^2 p_\mathrm {z}(\mathrm {z})d\mathrm {z}= 3 \mathbf {w}^{\scriptscriptstyle T}\Sigma _\mathbf {x}\mathbf {w}\mathcal{E}_{\mathrm {z}}\left\{ \mathrm {z}^2\right\} , \end{aligned}$$

where we use the fact that for a Gaussian distribution $\mathcal{N}_{t}(0,\sigma ^2)$, its fourth order moment is $3\sigma ^4$. Putting all results together in the definition of kurtosis, we have

Appendix 3: Derivation of Eq. (11)

First we expand the objective function in (7) as

$$\begin{aligned} L\left( \sqrt{\kappa },\sigma ^2\right) = \sum _{k=1}^K \left( \sqrt{\tilde{\kappa }_k} - \sqrt{\kappa } + \frac{\sqrt{\kappa }\sigma ^2}{{\tilde{\sigma }^2_{k}}}\right) ^2, \end{aligned}$$

The gradient of $L\left( \sqrt{\kappa },\sigma ^2\right)$ with regards to the two parameters are computed as, as:

$$\begin{aligned} \frac{\partial L}{\partial \sigma ^2}&= 2 \sum _{k=1}^K \left( \sqrt{\tilde{\kappa }_k} - \sqrt{\kappa } + \frac{\sqrt{\kappa }\sigma ^2}{{\tilde{\sigma }^2_{k}}} \right) \frac{\sqrt{\kappa }}{{\tilde{\sigma }^2_{k}}}.\end{aligned}$$

(18)

$$\begin{aligned} \frac{\partial L}{\partial \sqrt{\kappa }}&= 2 \sum _{k=1}^K \left( \sqrt{\tilde{\kappa }_k} - \sqrt{\kappa } + \frac{ \sqrt{\kappa }\sigma ^2}{{\tilde{\sigma }^2_{k}}}\right) \left( \frac{\sigma ^2}{{\tilde{\sigma }^2_{k}}} - 1\right) \end{aligned}$$

(19)

Setting Eq. (18) to zero, and considering $\sqrt{\kappa } > 0$, we have

$$\begin{aligned} \sum _{k=1}^K \frac{1}{{\tilde{\sigma }^2_{k}}} \left( \sqrt{\tilde{\kappa }_k} - \sqrt{\kappa } + \frac{\sqrt{\kappa }\sigma ^2}{{\tilde{\sigma }^2_{k}}} \right) = 0, \end{aligned}$$

(20)

Setting Eq. (19) to zero and substituting with Eq. (20) yield

$$\begin{aligned} \sum _{k=1}^K \left( \sqrt{\tilde{\kappa }_k} - \sqrt{\kappa } + \frac{\sqrt{\kappa }\sigma ^2}{{\tilde{\sigma }^2_{k}}} \right) = 0, \end{aligned}$$

from which we can obtain

$$\begin{aligned} \sigma ^2 = \frac{1}{\frac{1}{K}\sum _{k=1}^K \frac{1}{{\tilde{\sigma }^2_{k}}}} - \frac{1}{\sqrt{\kappa }} \frac{\sum _{k=1}^K \sqrt{\tilde{\kappa }_k}}{\sum _{k=1}^K \frac{1}{{\tilde{\sigma }^2_{k}}}}. \end{aligned}$$

(21)

Next, substituting Eq. (21) back into Eq. (20), we have

$$\begin{aligned}&\!\!\!\!\sqrt{\kappa } \left( \frac{1}{\frac{1}{K}\sum _{k=1}^K \frac{1}{{\tilde{\sigma }^2_{k}}}} - \frac{1}{\sqrt{\kappa }} \frac{\sum _{k=1}^K \sqrt{\tilde{\kappa }_k}}{\sum _{k=1}^K \frac{1}{{\tilde{\sigma }^2_{k}}}}\right) \sum _{k=1}^K \frac{1}{(\tilde{\sigma }^2_{k})^2}\nonumber \\&\!\!\! \quad +\sum _{k=1}^K \frac{\sqrt{\tilde{\kappa }_k}}{{\tilde{\sigma }^2_{k}}} - \sqrt{\kappa } \sum _{k=1}^K \frac{1}{{\tilde{\sigma }^2_{k}}} = 0. \end{aligned}$$

(22)

Further arranging terms and replacing average over different channels with $\langle \cdot \rangle _k$ yield Eq. (11). Further checking the second-order conditions ensures that the solution is the unique global minimizer of Eq. (7).