Show that every random variable $X$ can be written as $$X=\lambda Z_1+(1-\lambda)Z_2$$ for a discrete random variable $Z_1$, a continuous random variable $Z_2$, and a real value $\lambda$.

This exercise was proposed in class. I really don't know if the result is true.

I know the trivial cases ($X$ discrete and $X$ continuous), but need help in the general case. Can anybody help me?

  • $\begingroup$ Are you sure the "mixing" is done to the values of the random variables (and not to the densities or distribution functions)? That is, $X$ equals $Z_1$ with probablity $\lambda$ or $Z_2$ with prob $1-\lambda$ As stated, it doesn't make much sense to me. $\endgroup$ – leonbloy Aug 20 '15 at 20:07
  • $\begingroup$ The statement is equivalent to every nondiscrete random variable being $X=Y+Z$ for some discrete random variable $Y$ and some continuous random variable $Z$. How to do this, say, if $P_X$ is the midpoint of the uniform distribution on $(0,1)$ and the Dirac measure at $2$? $\endgroup$ – Did Aug 20 '15 at 20:18
  • $\begingroup$ Isn't the sum of a continuous random variable and a discrete random value also continuous? Am I missing something? $\endgroup$ – user251257 Aug 20 '15 at 21:10
  • $\begingroup$ @user251257 If they are independent, yes, and possibly in more cases. But how is this relevant? $\endgroup$ – Did Aug 21 '15 at 8:53
  • $\begingroup$ @Did I was wondering could $X$ be anything other then discrete or continuous. $\endgroup$ – user251257 Aug 21 '15 at 9:33

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