Contaminated normal distribution

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The contaminated normal distribution is a special form of mixed distribution . It plays a major role in robustness studies of estimators and tests .

The real random variable has a contaminated normal distribution if its density function is in the form ${\displaystyle X\,}$

${\displaystyle f(x)=(1-\varepsilon ){\frac {1}{{\sqrt {2\pi }}\sigma _{1}}}\operatorname {e} ^{-{\frac {1}{2}}\left({\frac {x-\mu _{1}}{\sigma _{1}}}\right)^{2}}+\varepsilon {\frac {1}{{\sqrt {2\pi }}\sigma _{2}}}\operatorname {e} ^{-{\frac {1}{2}}\left({\frac {x-\mu _{2}}{\sigma _{2}}}\right)^{2}}}$

with , i.e. as a convex combination of two normal distribution density functions. ${\displaystyle \varepsilon \in [0,1]}$

The distribution function then has the form

${\displaystyle F=(1-\varepsilon )\operatorname {N} (\mu _{1},\sigma _{1}^{2})+\varepsilon \operatorname {N} (\mu _{2},\sigma _{2}^{2})}$.

The distribution function of a normally distributed random variable applies here . ${\displaystyle \mu _{1},\mu _{2}\in \mathbb {R} ,\sigma _{1},\sigma _{2}>0,\operatorname {N} (\mu ,\sigma ^{2})}$

The following applies to the expected value and the variance :

${\displaystyle \operatorname {E} (X)=(1-\varepsilon )\mu _{1}+\varepsilon \mu _{2}}$,

${\displaystyle \operatorname {Var} (X)=(1-\varepsilon )\sigma _{1}^{2}+\varepsilon \sigma _{2}^{2}+\varepsilon (1-\varepsilon )(\mu _{1}-\mu _{2})^{2}}$.

Often, through additional conditions, such as special cases are derived ( scale-contaminated normal distribution ). ${\displaystyle \mu _{1}=\mu _{2}\,}$

example

A manufacturer of electronic devices uses capacitors with a capacity of 5 nF, which it purchases from two manufacturers. Those manufactured by A show a slightly smaller scatter than those from B. 60% of the capacitors purchased come from manufacturer A, 40% from B. Assume that the capacitance of the capacitors from both manufacturers is normally distributed with parameters in a sufficiently wide range . Be and . ${\displaystyle \mu _{1},\mu _{2},\sigma _{1}^{2},\sigma _{2}^{2}}$ ${\displaystyle \mu _{1}=\mu _{2}=5\,\operatorname {nF} \,}$ ${\displaystyle \sigma _{1}^{2}=0{,}0144\,\operatorname {nF} ^{2},\sigma _{2}^{2}=0{,}0225\,\operatorname {nF} ^{2}}$

A deviation of more than 10% from the nominal value of the capacitance is highly undesirable. What is the probability that a capacitor has a capacitance that differs by more than 10%?

${\displaystyle 10\,\%\,\cdot \,5=0{,}5\,,}$

${\displaystyle {\begin{aligned}P(X<4{,}5\cup X>5{,}5)&=\left(0{,}6\,\cdot \,\Phi \left({\frac {(5-0{,}5)-5}{\sqrt {0{,}0144}}}\right)\,+\,0{,}4\,\cdot \,\Phi \left({\frac {(5-0{,}5)-5}{\sqrt {0{,}0225}}}\right)\right)\\&\qquad +\left(1-0{,}6\,\cdot \,\Phi \left({\frac {(5+0{,}5)-5}{\sqrt {0{,}0144}}}\right)\,+\,1-0{,}4\,\cdot \,\Phi \left({\frac {(5+0{,}5)-5}{\sqrt {0{,}0225}}}\right)\right)\\&=\left(0{,}6\,\cdot \,\Phi \left({\frac {-0{,}5}{0{,}12}}\right)\,+\,0{,}4\,\cdot \,\Phi \left({\frac {-0{,}5}{0{,}15}}\right)\right)+\left(1-0{,}6\,\cdot \,\Phi \left({\frac {0{,}5}{0{,}12}}\right)\,+\,1-0{,}4\,\cdot \,\Phi \left({\frac {0{,}5}{0{,}15}}\right)\right)\\&=2\,\cdot \,\left(0{,}6\,\cdot \,\Phi \left({\frac {-0{,}5}{0{,}12}}\right)\,+\,0{,}4\,\cdot \,\Phi \left({\frac {-0{,}5}{0{,}15}}\right)\right)\\&\approx 2\,\cdot \,\left(0{,}6\,\cdot \,0{,}000015464+0{,}4\,\cdot \,0{,}000429117\right)\\&=0{,}000361849\end{aligned}}}$

A proportion of around 0.000361849 of all capacitors shows a deviation of more than 10% in terms of capacity.