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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

Thanks in advance.

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  • $\begingroup$ 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
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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.

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$Z$ 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.

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  • $\begingroup$ 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

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