# Whether the product of a continuous and a discrete random variables is continuous or discrete?

I have one Random Variable $$X$$ which is continuous and uniform in $$X\in [0,x_m]$$. Also, I have another Random Variable $$Y$$ which is discrete and uniform where $$Y \in \{-1,0,1\}$$ . So, the product say $$Z=XY$$ will be continuous or discrete random variable? Please explain in detail.

Additional info: $$X$$ and $$Y$$ are independent random variables.

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I solved for the PDF of $$Z$$ in the following way,

\begin{align*} P(Z\leq z) &= P(XY \leq z)\\ &=P(X \leq \frac{z}{Y})\\ &= P(X \leq \frac{z}{Y} \mid Y = -1)P(Y=-1) + P(X \leq \frac{z}{Y} \mid Y = 0)P(Y=0) + P(X \leq \frac{z}{Y} \mid Y=1) P(Y=1)\\ &=\frac{1}{3}P(-X \leq z) + \frac{1}{3}P(X.0 \leq z) + \frac{1}{3}P(X \leq z)\\ &=\frac{1}{3}P(X \geq -z) + \frac{1}{3}P(X < \infty) + \frac{1}{3}P(X \leq z)\\ &=\dfrac{1}{3}\int_{-z}^\infty \frac{1}{x_m}dx + \frac{1}{3}\int_{-\infty}^\infty\frac{1}{x_m}dx +\int_{-\infty}^z \frac{1}{x_m}dx\\ &=\dfrac{1}{3}\int_{-z}^{x_m} \frac{1}{x_m}dx + \dfrac{1}{3}.1 +\dfrac{1}{3}\int_{0}^z \frac{1}{x_m}dx\\ &=\frac{1}{3}\bigl(1+\frac{z}{x_m}\bigr) + \frac{1}{3} + \frac{1}{3}\dfrac{z}{x_m}\\ &=\frac{2}{3}\bigl(1+\frac{z}{x_m}\bigr) \end{align*}

So, the CDF of $$Z$$ is $$F_Z(z) = \dfrac{2}{3}\bigl(1+\frac{z}{x_m}\bigr)$$ from which the PDF is calculated as

$$f_Z(z) = \dfrac{d}{dz}F_Z(z)\\f_Z(z) = \dfrac{2}{3x_m}$$

But, I couldn't figure out whether I have done right or wrong and please solve for the PDF of Z if I am wrong and also mention it's ranges accordingly

• Are the random variable independent? What is your definition of continuous random variable? (There are conflicting definitions). – Kavi Rama Murthy Oct 26 '18 at 8:15
• Yes the random variables are independent and as the continuous random variable is uniform, it's PDF will be $1/x_m$ @KaviRamaMurthy – kunarapu priyatham Oct 26 '18 at 8:17
• @kunarapupriyatham Such essential info needs a place in your question. A comment is not enough. – drhab Oct 26 '18 at 8:34
• Thank you @drhab for guiding me as I am new to the community. – kunarapu priyatham Oct 26 '18 at 8:37
• @kunarapupriyatham Next time: pose a new question if you ask for something new. This eventually with a link to a question that was firstly posted and is connected to the new question. – drhab Oct 26 '18 at 12:43

We have $$P\{XY=0\} \geq P\{Y=0\} >0$$ so $$XY$$ is not continuous. It is not discrete either. If it is discrete it can take only a counatble num ber of values $$c_1,c_2,\cdots$$. Split $$1=P\{XY\in \{c_1,c_2,\cdots\}\}$$ into the parts $$Y=0$$, $$Y=1$$ and $$Y=-1$$ to get a contradiction.

$$Z$$ is a continuous rv iff for every $$z\in\mathbb R$$ we have $$P(Z=z)=0$$

Now observe that $$P(Z=z)=$$$$P(Z=z\mid Y=-1)P(Y=-1)+P(Z=z\mid Y=0)P(Y=0)+P(Z=z\mid Y=1)P(Y=1)=$$$$P(X=-z)P(Y=-1)+P(0=-z)P(Y=0)+P(X=z)P(Y=1)=$$$$P(0=-z)P(Y=0)$$

This makes clear that $$P(Z=z)=0$$ for every $$z\neq0$$, but also that $$P(Z=0)=P(Y=0)$$.

Our conclusion is that $$Z$$ is a continuous rv iff $$P(Y=0)=0$$.

Since $$Y$$ has uniform distribution over $$\{-1,0,1\}$$ this is not the case.

Final conclusion: $$Z$$ is not a continuous random variable.

Edit (concerning newly inserted question.

Calculation of CDF of $$Z$$ goes like this:

\begin{aligned}F_{Z}\left(z\right) & =P\left(XY\leq z\mid Y=-1\right)P\left(Y=-1\right)+P\left(XY\leq z\mid Y=0\right)P\left(Y=0\right)+P\left(XY\leq z\mid Y=1\right)P\left(Y=1\right)\\ & =P\left(-X\leq z\right)\frac{1}{3}+P\left(0\leq z\right)\frac{1}{3}+P\left(X\leq z\right)\frac{1}{3}\\ & =\frac{1}{3}\left[P\left(X\geq-z\right)+\mathbf{1}_{\left[0,\infty\right)}\left(z\right)+P\left(X\leq z\right)\right] \end{aligned}

Now we discern the following cases:

\begin{aligned}\bullet\; & z<-x_{m} & \text{gives } & F_{Z}\left(z\right)=0\\ \bullet\; & -x_{m}\leq z<0 & \text{gives } & F_{Z}\left(z\right)=\frac{z+x_{m}}{3x_{m}}\\ \bullet\; & 0\leq z\leq x_{m} & \text{gives } & F_{Z}\left(z\right)=\frac{z+2x_{m}}{3x_{m}}\\ \bullet\; & z>x_{m} & \text{gives } & F_{Z}\left(z\right)=1 \end{aligned}

For the existence of a PDF it is necessary (not sufficient) that $$Z$$ is a continuous random variable.

As made clear in the answer of Kavi this is not the case so $$Z$$ has no PDF.

For the existence of a PMF it is necessary (and sufficient) that $$Z$$ is a discrete random variable.

As made clear in the answer of Kavi this is not the case so $$Z$$ has no PMF.

We can at most write $$F_Z(z)=\frac23G(z)+\frac13H(z)$$ where:

• $$H(z):=\mathsf1_{[0,\infty)}(z)$$ is the CDF of a discrete random variable with PMF defined by $$x\mapsto1$$ if $$x=0$$ and $$x\mapsto0$$ otherwise.

• $$G(z)$$ is defined by $$x\mapsto0$$ if $$z<-x_m$$, $$x\mapsto\frac{z+x_m}{2x_m}$$ if $$-x_m\leq z\leq x_m$$ and $$x\mapsto1$$ otherwise. It has a PDF prescribed by $$x\mapsto\frac1{2x_m}$$ if $$-x_m\leq z\leq x_m$$ and $$x\mapsto0$$ otherwise.

• The answer of @Kavi is more complete. He also shows that $Z$ is not discrete. Feel free to change your acceptance. – drhab Oct 26 '18 at 8:41
• It is nice of you to make this comment. Thanks! – Kavi Rama Murthy Oct 26 '18 at 8:44