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I have a random variable $x_1$ that follows the normal distribution with mean 0 and variance 1 with probability 0.6 and follows the normal distribution with mean 0 and variance 2 with probability 0.4. How can I write this in a mathematical way so I can calculate the pdf of $x_1$ and find the joint pdf with another random variable?

Thank you.

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Let $Y$ a Bernoulli distributed random variable with parameter $p=0.6$, that is $P(Y=1)=1-P(Y=0)=0.6$.

Then you have the conditional distribution of $X_1$ given $Y=1$:

$$X_1\mid Y=1\sim\mathcal{N}(0,1)$$ Considering the pdf: $$f_{X_1\mid Y=1}(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$ You also have: $$X_1\mid Y=0\sim\mathcal{N}(0,2)$$ and $$f_{X_1\mid Y=0}(x)=\frac{1}{2\sqrt{\pi}}e^{-\frac{x^2}{4}}$$

Finally, the pdf of $X_1$ is given by: $$f_{X_1}(x)=P(Y=1)f_{X_1\mid Y=1}(x)+P(Y=0)f_{X_1\mid Y=0}(x)= 0.6\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}+0.4\frac{1}{2\sqrt{\pi}}e^{-\frac{x^2}{4}}$$

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$$ \Pr(X\le x) = 0.6 \Phi(x) + 0.4 \Phi\left( \frac x {\sqrt 2} \right) $$ $$ \frac d {dx} \Pr(X\le x) = 0.6 \varphi(x) + 0.4 \varphi\left( \frac x {\sqrt 2} \right) \cdot \frac 1 {\sqrt 2} $$ (The chain rule was used in the second term.)

Recall that $\displaystyle \varphi(x) = \frac 1 {\sqrt{2\pi}} e^{-x^2/2}$.

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