Welcome to our community

Be a part of something great, join today!

Distribution and Density functions of maximum of random variables

Dhamnekar Winod

Active member
Nov 17, 2018
100
1] Let X,Y,Z be independent, identically distributed random variables, each with density $f(x)=6x^5$ for $0\leq x\leq 1,$ and 0 elsewhere. How to find the distributon and density functions of the maximum of X,Y,Z.


2]Let X and Y be independent random variables, each with density $e^{-x},x\geq 0$.How to find the distribution and density functions of $Z=\frac{Y}{X}?$

At present, I am studying CDF, PDF and MGF techniques for transformations of random variables. I am searching for answers for similar types of questions on internet. Meanwhile if any member knows correct answers, may reply with correct answers.

I have computed pdf for $Z=\frac{Y}{X}$ as $\frac{e^{(-2x)}}{x},x>0$. Is that correct?
 
Last edited:

steep

Member
Dec 17, 2018
51
1] Let X,Y,Z be independent, identically distributed random variables, each with density $f(x)=6x^5$ for $0\leq x\leq 1,$ and 0 elsewhere. How to find the distributon and density functions of the maximum of X,Y,Z.


2]Let X and Y be independent random variables, each with density $e^{-x},x\geq 0$.How to find the distribution and density functions of $Z=\frac{Y}{X}?$

At present, I am studying CDF, PDF and MGF techniques for transformations of random variables. I am searching for answers for similar types of questions on internet. Meanwhile if any member knows correct answers, may reply with correct answers.

I have computed pdf for $Z=\frac{Y}{X}$ as $\frac{e^{(-2x)}}{x},x\geq 0$. Is that correct?
do these one at a time. You should show your work so we can follow what you know /don't know.

In both cases you want to work with CDFs not PDFs as much as possible. For the first question consider the event
$\{X\leq t\} \cap \{Y\leq t\} \cap \{Z\leq t\}$
how would you compute that, and what does it have to do with the CDF for the maximal random variable?

Once you have the desired CDF, just differentiate at the end to recover the PDF.
 

Dhamnekar Winod

Active member
Nov 17, 2018
100
do these one at a time. You should show your work so we can follow what you know /don't know.

In both cases you want to work with CDFs not PDFs as much as possible. For the first question consider the event
$\{X\leq t\} \cap \{Y\leq t\} \cap \{Z\leq t\}$
how would you compute that, and what does it have to do with the CDF for the maximal random variable?

Once you have the desired CDF, just differentiate at the end to recover the PDF.
Hello,
Answer to 2) question is as follows:

The joint density of X and Y is given by $f_{X,Y}(x,y)=e^{-x}*e^{-x}=e^{-2x}$

Now if $g_1(x,y)=\frac{y}{x}, g_2(x,y)=x,$ then

$\frac{\partial{g_1}}{\partial{x}}=y$, $\frac{\partial{g_1}}{\partial{y}}=\frac{1}{x}$

$\frac{\partial{g_2}}{\partial{x}}=1$, $\frac{\partial{g_2}}{\partial{y}}=0$

and so,

$$J(x,y)=\left| \begin{matrix}y & \frac{1}{x}\\ 1&0\end{matrix} \right|=\frac{-1}{x}$$

Finally, $z=\frac{y}{x}, v=x$ have as their solutions x=v, y=zv, we see that

$$f_{V,Z}(v,z)=f_{x,y}[v,zv]v=ve^{-2v}$$

Now$$f_Z(z)=\displaystyle\int_0^\infty x*e^{-2x}dx=0.25$$

Part 2: Answer to 1) the event $(X\leq t)\cap(Y\leq t)\cap(Z\leq t)=1$. But how can we compute PDF and CDF of maximum of X,Y,Z.
 

steep

Member
Dec 17, 2018
51
Part 2: Answer to 1) the event $(X\leq t)\cap(Y\leq t)\cap(Z\leq t)=1$. But how can we compute PDF and CDF of maximum of X,Y,Z.
No. The probability of those 3 events occurring is not in general 1. A nit pick but an important one: the intersection of 3 events does not spit out a number either -- it's only after you apply a probability function to a collection of events-- that maps the underlying sample points to a number in $\in [0,1]$

I am telling you the intersection of those 3 events has something to do with the CDF of the maximum of those 3 random variables...
= = = = =
Based on this response: I'd strongly suggest consulting an outside text as there are material gaps that need addressed. The subject matter you're dealing with in particular is called Order Statistics. They are nicely treated in Blitzstein and Hwang's book, which is now freely available here:

https://projects.iq.harvard.edu/stat110/home
 

Dhamnekar Winod

Active member
Nov 17, 2018
100
Hello,
I have computed CDF of maximum of $X,Y,Z$ i.i.d. random variables=$X^{18}$ and its PDF is =$18*X^{17}$.Are my computation correct?
 

steep

Member
Dec 17, 2018
51
Hello,
I have computed CDF of maximum of $X,Y,Z$ i.i.d. random variables=$X^{18}$ and its PDF is =$18*X^{17}$.Are my computation correct?
If I read between the lines, basically yes, though the way you've written it is confusing as capital $X$ is typically reserved for a random variable $X$, and of course you'd need to specify the domain

So the CDF of your original random variables is given by $F_X(t) = t^6$ for $t\in [0,1]$. They are iid so $F_Y(t) = t^6$ and $F_Z(t) = t^6$.

Since they are independent then
$P\big(\max\{X,Y,Z\} \leq t\big)= P\big((X\leq t)\cap(Y\leq t)\cap(Z\leq t)\big) = P(X\leq t)\cdot P(Y\leq t)\cdot P(Z\leq t)= F_X(t) \cdot F_Y(t) \cdot F_Z(t) = t^{18}$
for $t\in [0,1]$, and 1 for $t \gt 1$, and 0 for $t \lt 0$
 

Dhamnekar Winod

Active member
Nov 17, 2018
100
If I read between the lines, basically yes, though the way you've written it is confusing as capital $X$ is typically reserved for a random variable $X$, and of course you'd need to specify the domain

So the CDF of your original random variables is given by $F_X(t) = t^6$ for $t\in [0,1]$. They are iid so $F_Y(t) = t^6$ and $F_Z(t) = t^6$.

Since they are independent then
$P\big(\max\{X,Y,Z\} \leq t\big)= P\big((X\leq t)\cap(Y\leq t)\cap(Z\leq t)\big) = P(X\leq t)\cdot P(Y\leq t)\cdot P(Z\leq t)= F_X(t) \cdot F_Y(t) \cdot F_Z(t) = t^{18}$
for $t\in [0,1]$, and 1 for $t \gt 1$, and 0 for $t \lt 0$
Hello,
$P\big(\min\{X,Y,Z\}\leq t\big)=1-(1-t^6)^3$ and its density is $18*t^5(1-t^6)^2$, for$ t\in [0,1]$ and 1for$ t\gt 1$ and 0 for $t \lt 0$