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Tuorui "v1ncent19" Peng

En voyage dans l'espace de Hilbert.

Mathematics & Statistics2 min readEnglish

Variance of Odds Ratio in Contingency Table

Odds ratio of the contingency table below is an important statistics

θ=P(DE)/P(DE)P(DE)/P(DE)\begin{align} \theta =\dfrac{\mathbb{P}\left( D\vert E \right) \big/\mathbb{P}\left( D^\complement\vert E \right) }{\mathbb{P}\left( D\vert E^\complement\right)\big/\mathbb{P}\left( D^\complement \vert E^\complement \right) } \end{align}
$$D^{\,}$$ $$D^\complement$$
$$E$$ $$n_{11} $$$$n_{12} $$
$$E^\complement$$$$n_{21}$$$$n_{22}$$

Estimation of odds ratio is symmetric w.r.t. retrospective/prospective study:

θ^=n11n22n12n21\begin{align} \hat{\theta }=\dfrac{n_{11}n_{22}}{n_{12}n_{21}} \end{align}

its variance is derived with assumption (n11,n12,n21,n22)M(p11,p12,p21,p22;n)(n_{11},n_{12},n_{21},n_{22})\sim M(p_{11},p_{12},p_{21},p_{22}; n)

Covariance of Multinoimal Distribution

for a1,,anM(p1,,pn;a)a_1,\ldots,a_n\sim M(p_1,\ldots,p_n;a):

cov(ai,aj)=E((aiapi)(ajapj))=E(aiaj)a2pipj=E(aiE(ajai))a2pipj=E(ai(aai)pj1pi)a2pipj=pj1piE(aaiai2)a2pipj=pj1pi[a2pi(api(1pi)+a2pi2)]a2pipj=apipj,ij\begin{align} cov(a_i,a_j)=&\mathbb{E}\left( (a_i-ap_i)(a_j-ap_j) \right)\\ =&\mathbb{E}\left( a_ia_j \right)-a^2p_ip_j\\ =&\mathbb{E}\left( a_i\mathbb{E}\left( a_j\vert a_i \right) \right) -a^2p_ip_j\\ =&\mathbb{E}\left( a_i (a-a_i)\dfrac{p_j}{1-p_i} \right) -a^2p_ip_j\\ =&\dfrac{p_j}{1-p_i} \mathbb{E}\left( aa_i-a_i^2 \right) -a^2p_ip_j\\ =&\dfrac{p_j}{1-p_i}\left[ a^2p_i-\left( ap_i(1-p_i) + a^2p_i^2 \right) \right] -a^2p_ip_j\\ =&-ap_ip_j,\quad i\neq j \end{align}

i.e.

cov(ai,aj)={apipj,ijapi(1pi),i=j=a(δijpipipj)\begin{align} cov(a_i,a_j)=\begin{cases} -ap_ip_j,&i\neq j\\ ap_i(1-p_i),&i=j \end{cases}=a\left(\delta _{ij}p_i-p_ip_j\right) \end{align}

Delta Method

To estimate variance of function of r.v., say var(f(X))var(f(X)), use the taylor expansion of f()f(\, \cdot \, ) at E(X)\mathbb{E}(X):

f(X)f(E(X))+f(E(X))(XE(X))+O(f)\begin{align} f(X)\approx f\left(\mathbb{E}\left( X \right) \right)+\nabla f(\mathbb{E}\left( X \right) )'\left(X-\mathbb{E}\left( X \right) \right)+\mathcal{O}(\nabla\nabla f) \end{align}

then

var(f(X))(f(E(X)))var(X)(f(E(X)))\begin{align} var(f(X))\approx & \left(\nabla f(\mathbb{E}\left( X \right) )\right)' var(X) \left(\nabla f(\mathbb{E}\left( X \right) )\right) \end{align}

Similarly for bi-function

then

cov(f(X),g(Y))(xf)cov(X,Y)(yg)\begin{align} cov(f(X),g(Y))\approx & \left(\nabla_x f\right)' cov(X,Y) \left(\nabla_y g\right) \end{align}

using delta method for multinomial distribution, we could obtain that

cov(logai,logaj)=1a2pipjcov(ai,aj)=(δijpipipj)apipj=δijni1a\begin{align} cov(\log a_i,\log a_j)=&\dfrac{1}{a^2p_ip_j}cov(a_i,a_j)=\dfrac{\left(\delta _{ij}p_i-p_ip_j\right)}{ap_ip_j}=\dfrac{\delta _{ij}}{n_i}-\dfrac{1}{a} \end{align}

Variance of Odds Ratio

var(logn11n22n12n21)=var(logn11+logn22log21log12)=(ij)var(lognij)+2(ij,kl){(11,22),(12,21)}cov(lognij,lognkl)2(ij,kl){(11,22),(12,21)}cov(lognij,lognkl)=(ij)1nij4n4n+8n=1n11+1n21+1n12+1n22\begin{align} var\left( \log \dfrac{n_{11}n_{22}}{n_{12}n_{21}} \right)=&var\left( \log n_{11}+\log n_{22}-\log_{21}-\log_{12} \right)\\ =&\sum_{(ij)}var(\log n_{ij})+2\sum_{(ij,kl)\in \{(11,22),(12,21)\}} cov(\log n_{ij},\log {n_{kl}})\\ &-2\sum_{(ij,kl)\notin \{(11,22),(12,21)\}} cov(\log n_{ij},\log {n_{kl}})\\ =&\sum_{(ij)}\dfrac{1}{n_{ij}}-\dfrac{4}{n}-\dfrac{4}{n}+\dfrac{8}{n}\\ =&\dfrac{1}{n_{11}}+\dfrac{1}{n_{21}}+\dfrac{1}{n_{12}}+\dfrac{1}{n_{22}} \end{align}

And use delta method again to obtain variance for odds ratio

var(n11n22n12n21)=(n11n22n12n21)2(1n11+1n21+1n12+1n22)\begin{align} var(\dfrac{n_{11}n_{22}}{n_{12}n_{21}})=\left(\dfrac{n_{11}n_{22}}{n_{12}n_{21}}\right)^2\left(\dfrac{1}{n_{11}}+\dfrac{1}{n_{21}}+\dfrac{1}{n_{12}}+\dfrac{1}{n_{22}}\right) \end{align}

Comment: the above could be used to multi-row/column contingency table on four aligned grids, e.g.

var(n22n44n24n42)=(n22n44n24n42)2(1n22+1n44+1n24+1n42)\begin{align} var(\dfrac{n_{22}n_{44}}{n_{24}n_{42}})=\left(\dfrac{n_{22}n_{44}}{n_{24}n_{42}}\right)^2\left(\dfrac{1}{n_{22}}+\dfrac{1}{n_{44}}+\dfrac{1}{n_{24}}+\dfrac{1}{n_{42}}\right) \end{align}