Consider the following probability distribution function (PDF):

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

$p(x)$ 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 $p(x)$.

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    $\begingroup$ 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. $\endgroup$
    – Glen_b
    Commented Feb 3 at 2:33
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    $\begingroup$ Thank you as soon as I realized this can be searched as a mixture distribution I was good! $\endgroup$ Commented Feb 3 at 2:46

1 Answer 1


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, $$ \P(Z \le z) = \\ \P(Z \le z \mid \text{coin}=F) \cdot \P(\text{coin}=F) + \P(Z \le z \mid \text{coin}=G) \cdot \P(\text{coin}=G) = \\ F(z)\cdot a + G(z)\cdot b $$

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    $\begingroup$ 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. $\endgroup$
    – Xi'an
    Commented Feb 3 at 14:32
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    $\begingroup$ 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. $\endgroup$ Commented Feb 4 at 5:08

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