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Consider two independent random variables $X$ and $Y$, where $X$ is uniformly distributed on the interval $[0,1]$ and $Y$ is uniformly distributed on the set $\{0,1\}$. Thus, the cdfs are given by $F_X(x) = \begin{cases} 0 & x <0\\ x & 0 \leq x <1\\ 1 & else \end{cases}$

and

$F_Y(y) = \begin{cases} 0 & y <0\\ 1/2 & 0 \leq y <1\\ 1 & else. \end{cases}$

Consider the random variable $Z = (X,Y)$ with cdf is given by

$F_Z(x,y) = \begin{cases} 0 & y<0\\ F_X(x)/2& 0 \leq y < 1\\ F_X(x) & else. \end{cases}$

Now, what I'm interested in is a pdf of $Z$. Is

$ f_Z(x,y) = \begin{cases} 0 & y<0\\ f_X(x)/2& 0 \leq y < 1\\ f_X(x) & else. \end{cases} $

the pdf of $Z$? More generally, I'm interested in the joint pdf of independent random variables, one of which is continuous and the others (possibly more than one) are discrete. If correct, can the above be applied in this case?

Thank you for your help.

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1 Answer 1

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No. If one of the variables is discrete and the other continuous, they can't have a common density [neither with respect to the Lebesgue-measure, nor the counting measure].

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  • $\begingroup$ Ah, ok, I see. Thanks! $\endgroup$ Commented Sep 23, 2015 at 16:02
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    $\begingroup$ The expression is simply wrong. They can't have a joint density, since the set $N = \mathbb{R} \times \{0, 1\}$ is a Lebesgue-Nullset, but $P(Z \in N) = 1$. $\endgroup$ Commented Sep 23, 2015 at 19:55
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    $\begingroup$ Yes, that's because the random variables $X$ and $Y$ are independent. This implies $P(Z \in [1, 1/2] \times \{0\}) = P(X \in [0, 1/2], Y \in \{0\}) = P(X \in [0, 1/2])P(Y \in \{0\}) = 1/4$. $\endgroup$ Commented Sep 23, 2015 at 20:08
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    $\begingroup$ Yes, this doesn't even have anything to do with the independence. The join distribution of a continuous and a discrete random variable never has a pdf. $\endgroup$ Commented Sep 23, 2015 at 20:15
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    $\begingroup$ No, the set $N$ has measure $1$ with respect to measure that is induced by $Z$. The problem isn't any kind of definition of the distribution of $Z$ [because its distribution is well-defined], you just can't write a density of this distribution with respect to either the Lebesgue- or counting measure. To see that it is true for the counting measure, observe that $P(Z \in \{(a, b)\}) = 0$ for all $a, b \in \mathbb{R}$. $\endgroup$ Commented Sep 23, 2015 at 20:28

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