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.

----------------------------------Newly Inserted-------------------------------

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

Thanks in advance.

  • $\begingroup$ Are the random variable independent? What is your definition of continuous random variable? (There are conflicting definitions). $\endgroup$ – Kavi Rama Murthy Oct 26 '18 at 8:15
  • $\begingroup$ Yes the random variables are independent and as the continuous random variable is uniform, it's PDF will be $1/x_m$ @KaviRamaMurthy $\endgroup$ – kunarapu priyatham Oct 26 '18 at 8:17
  • $\begingroup$ @kunarapupriyatham Such essential info needs a place in your question. A comment is not enough. $\endgroup$ – drhab Oct 26 '18 at 8:34
  • $\begingroup$ Thank you @drhab for guiding me as I am new to the community. $\endgroup$ – kunarapu priyatham Oct 26 '18 at 8:37
  • $\begingroup$ @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. $\endgroup$ – 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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.