If $X$ is a discrete and $Y$ is a continuous random variable then what can we say about the distribution of $X+Y$? Is it continuous or is it mixed?
What about the product $XY$?
Solved – Is the sum of a discrete and a continuous random variable continuous or mixed
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Related Solutions
It's wrong because - as the answer explained - there are discrete atoms at 0 and 2.
By that cdf, you can wait exactly 0 time with positive probability (similarly with 2). Because of that, the waiting time is mixed, not continuous.
Presumably you've been given definitions of all three. How are continuous r.v.s defined?
If it's not immediately clear from the formula, it often helps to draw the cdf:
If you have a finite number of samples then your intuition is correct that the sample mean cannot take certain values. You mention that the discrete values are integers so clearly their mean cannot be irrational.
We know from the central limit theory that the sample mean tends to a continuous distribution as the sample size becomes large.
Even though a finite sample size doesn't technically create a continuous distribution, if your sample size is large or if the distance between discrete values is small for your purposes then you really should consider modelling it as a continuous variable.
Best Answer
Suppose $X$ assumes values $k \in K$ with discrete distribution $(p_k)_{k \in K}$, where $K$ is a countable set, and $Y$ assumes values in $\mathbb R$ with density $f_Y$ and CDF $F_Y$.
Let $Z = X + Y$. We have $$ \mathbb P( Z \leq z) = \mathbb P(X + Y \leq z) = \sum_{k \in K} \mathbb P(Y \leq z - X \mid X = k) \mathbb P(X = k) = \sum_{k \in K} F_Y(z-k) p_k,$$ which can be differentiated to obtain a density function for $Z$ given by $$ f_Z(z) = \sum_{k \in K} f_Y(z-k) p_k.$$
Now let $R = X Y$ and assume $p_0 = 0$. Then $$ \mathbb P(R \leq r) = \mathbb P(X Y \leq r) = \sum_{k \in K} \mathbb P(Y \leq r/X) \mathbb P(X= k) = \sum_{k \in K} F_Y(r/k) p_k,$$ which again can be differentiated to obtain a density function.
However if $p_0 > 0$, then $\mathbb P(X Y = 0) \geq \mathbb P(X = 0) = p_0 > 0$, which shows that in this case $XY$ has an atom at 0.