Template-Free Estimation of Intracranial Volume: A Preterm Birth Animal Model Study

Abstract

Accurate estimation of intracranial volume (ICV) is key in neuro-imaging-based volumetric studies, since estimation errors directly propagate to the ICV-corrected volumes used in subsequent analyses. ICV estimation through registration to a reference atlas has the advantage of not requiring manually delineated data, and can thus be applied to populations for which labeled data might be inexistent or scarce, e.g., preterm born animal models. However, such method is not robust, since the estimation depends on a single registration. Here we present a groupwise, template-free ICV estimation method that overcomes this limitation. The method quickly aligns pairs of images using linear registration at low resolution, and then computes the most likely ICV values using a Bayesian framework. The algorithm is robust against single registration errors, which are corrected by registrations to other subjects. The algorithm was evaluated on a pilot dataset of rabbit brain MRI ($N=7$), in which the estimated ICV was highly correlated ($\rho =0.99$) with ground truth values derived from manual delineations. Additional regression and discrimination experiments with human hippocampal volume on a subset of ADNI ($N=150$) yielded reduced sample sizes and increased classification accuracy, compared with using a reference atlas.

Appendix: Details of the Inference Algorithm

Replacing $p(\varvec{S} | c, \varvec{v})$ and $p(c | \alpha ,\beta )$ (Fig. 1b) in the first integral of Eq. 1:

$$\begin{aligned} \int _c (2c)^{-|\mathcal {S}|} \exp \left( - \frac{1}{c} \sum _{(i,j)\in \mathcal {S}} |S_{ij}-v_i+v_j| \right) \frac{\beta ^\alpha }{\varGamma (\alpha )} c^{-\alpha -1} \exp (-\beta /c) dc. \end{aligned}$$

(3)

Defining $\alpha ' = \alpha + |\mathcal {S}|$ and $\beta '=\beta +\sum _{(i,j)\in \mathcal {S}} |S_{ij}-v_i+v_j|$, Eq. 3 becomes:

$$\begin{aligned} \frac{1}{2^{|\mathcal {S}|}} \frac{\varGamma (\alpha ')}{\varGamma (\alpha )} \frac{\beta ^\alpha }{\beta '^{\alpha '}} \int _c \frac{\beta '^{\alpha '}}{\varGamma (\alpha ')} c^{-\alpha '-1} \exp (-\beta '/c) dc = \frac{1}{2^{|\mathcal {S}|}} \frac{\varGamma (\alpha ')}{\varGamma (\alpha )} \frac{\beta ^\alpha }{\beta '^{\alpha '}}, \end{aligned}$$

(4)

as the integral is over the probability density of $IG(\alpha ',\beta ')$ and thus equal to 1.

For the second integral in Eq. 1 (over $\mu$ and $\sigma ^2$), substitution of the expressions for the probabilities (again, see Fig. 1b in the paper) yields:

$$\begin{aligned} \int _\mu&\int _{\sigma ^2} \frac{1}{(2\pi \sigma ^2)^{N/2}} \exp \left( -\frac{1}{2\sigma ^2} \sum _{i=1}^N (v_i-\mu )^2 \right) \ldots \nonumber \\&\times \frac{b^{a}}{\varGamma \left( a\right) } \left( \sigma ^2\right) ^{-a-1} \exp \left( -b/\sigma ^2\right) \frac{\sqrt{n}}{\sqrt{2\pi \sigma ^2}} \exp \left[ -\frac{n}{2\sigma ^2}\left( \mu - m\right) ^2\right] d\mu d\sigma ^2. \end{aligned}$$

(5)

We now define $m'=(nm+N\bar{v})/(n+N)$, $n'=n+N$, $a'=a+\frac{N}{2}$, and:

$$ b' = b+\frac{1}{2} \sum _{i=1}^N (v_i-\bar{v})^2+\frac{nN}{n+N} \frac{(\bar{v}-m)^2}{2}, $$

where $\bar{v}$ is the average of $\varvec{v}$. Then, Eq. 5 becomes:

$$\begin{aligned} \int _\mu \int _{\sigma ^2}&\frac{b'^{a'}}{\varGamma \left( a'\right) } \left( \sigma ^2\right) ^{-a'-1} \exp \left( -b'/\sigma ^2\right) \frac{\sqrt{n'}}{\sqrt{2\pi \sigma ^2}} \exp \left[ -\frac{n'}{2\sigma ^2}\left( \mu - m'\right) ^2\right] d\mu d\sigma ^2 \ldots \nonumber \\&\times \frac{1}{(2\pi )^{N/2}} \sqrt{\frac{n}{n'}} \frac{b^a}{b'^{a'}} \frac{\varGamma (a')}{\varGamma (a)} = \frac{1}{(2\pi )^{N/2}} \sqrt{\frac{n}{n'}} \frac{b^a}{b'^{a'}} \frac{\varGamma (a')}{\varGamma (a)}, \end{aligned}$$

(6)

since the integral is over the probability density function of $NIG(m',n',a',b')$ and hence equal to 1.

Combining Eqs. 4 and 6, the problem in Eq. 1 becomes:

where z groups the terms independent of $\varvec{v}$. Taking the negated logarithm:

$$ \mathcal {C} = \alpha ' \log \beta ' + a' \log b' + \log z. $$

Substituting $a'$, $b'$, $\alpha '$ and $\beta '$ into this equation, and defining $Z=\log z$, we finally obtain the cost function in Eq. 2.