# linear Combination of Normal and T-Distributions

Consider the following probability distribution function (PDF):

$$$$p(x) = a\mathcal{N}(x; \mu, \sigma^2) + b \mathcal{T}(x; \mu, \tau^2, v) \; \; st. \; \;a + b = 1$$$$

$$p(x)$$ is clearly a symmetric PDF centered at $$\mu$$.

Is there any way to express the CDF of this function in terms of the CDF of the normal and $$t$$-distribution? Ideally I'd like to calculate the 95% confidence interval of $$p(x)$$.

• The cdfs mix exactly the same way. Proof is basic: differentiation of cdfs or integration of pdfs. It is also covered in the wikipedia article on mixture distributions. Commented Feb 3 at 2:33
• Thank you as soon as I realized this can be searched as a mixture distribution I was good! Commented Feb 3 at 2:46

What you have is a mixture distribution, and that the components of the mixture are normal and t is irrelevant. Let $$F$$ and $$G$$ be two cdfs (cumulative distribution functions) and consider the mixture distribution $$aF + bG$$ with your notation. Then, if $$Z$$ is a random variable distributed with this mixture, that is, $$Z \sim aF + bG$$, we have that $$\DeclareMathOperator{\P}{\mathbb{P}} \P(Z \le z)= (aF+bG)(z)= aF(z) + b G(z)$$ To see this, think about generating $$Z$$ as a mixture experiment. First, toss a coin, results are $$F$$ with probability $$a$$, $$G$$ with probability $$b$$. Then, conditional on the coin, generate $$Z$$ with distribution $$F$$ or $$G$$. Formally, $$\P(Z \le z) = \\ \P(Z \le z \mid \text{coin}=F) \cdot \P(\text{coin}=F) + \P(Z \le z \mid \text{coin}=G) \cdot \P(\text{coin}=G) = \\ F(z)\cdot a + G(z)\cdot b$$
• While the cdf is available as the linear combination of the two (Normal and t) cdfs, finding a 95% confidence interval$$(-\epsilon,\epsilon)$$requires solving (in $\epsilon$) the equation$$a\Phi(\sigma^{-1}\{\epsilon-\mu\})+bT_v(\tau^{-1}\{\epsilon-\mu\})=0.975$$which does not allow for a closed form expression. Commented Feb 3 at 14:32
• I assume that is the CDF of $T_v$? If so agreed, though with the CDF an educated staring guess and wandering parameter it should be pretty quick to approximate. Commented Feb 4 at 5:08