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Let $Z\sim \exp(1)$. Let $X$ be a new random variable (rv) defined as follows: We flip a coin. If we get head, than $X=Z$, and if we get tail than $X=-Z$.

I'm trying to figure whether $X$ is a discrete, continuous or mixed type rv, and to calculate its CDF and PDF (if it has one), but couldn't arrive at a solution.

Any help will be greatly appreciated.

Jimmy R.
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Brassican
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2 Answers2

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This is a mixture distribution $$F_X(x)=\frac12F_Z(x)+\frac12F_{-Z}(x)$$ for $x\in \mathbb R$, where the weights $1/2$ correspond to the result of the coin flip (assuming a fair coin, $1/2$ head and $1/2$ tail). Now $$F_{-Z}(x)=P(-Z\le x)=P(Z\ge -x)=1-F_Z(-x)$$ so that you can write $F_X(x)$ for $x\in \mathbb R$ as $$F_X(x)=\frac12F_Z(x)+\frac12\left(1-F_{Z}(-x)\right)$$ If you differentiate the previous equation you get the density of $X$ (so, yes, it has one) $$f_X(x)=\frac12 f_Z(x)+\frac12f_Z(-x)$$ for $x\in \mathbb R$.

Jimmy R.
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  • Thanks. It's not clear to me why you say that it's a mixture distribution. We get $F_X(x)=\frac12(1+e^{(-x)}-e^x)$, which is a continuous PDF, so why $X$ is not a continuous rv? – Brassican Dec 17 '15 at 09:42
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    Yes, $X$ is continuous as a mixture of continuous random variables. It seems that you ignored the link I sent you. Now, be careful, the $F$ you have is false. $F_Z$ changes its type when $x<0$ and $x>0$. For example when $x>0$ then $F(x)=1-e^{-x}$ but $F(-x)=0$!! – Jimmy R. Dec 17 '15 at 09:46
  • I get it now. I thought earlier that by definition, a mixture distribution must be a mixture of both a discrete rv and a continuous rv. – Brassican Dec 17 '15 at 13:27
  • @Brassican Ok, yes, this can be confusing indeed. – Jimmy R. Dec 17 '15 at 13:55
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Preassumed that the coin is fair:

$$F_{X}\left(x\right)=P\left(X\leq x\right)=P\left(X\leq x\mid\text{Head}\right)P\left(\text{Head}\right)+P\left(X\leq x\mid\text{Tail}\right)P\left(\text{Tail}\right)=P\left(Z\leq x\right)\frac{1}{2}+P\left(-Z\leq x\right)\frac{1}{2}$$

et cetera.

drhab
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