Distribution and Density functions of maximum of random variables

In summary: Therefore the CDF of the maximum of $X,Y,Z$ is given by$$F_{\max\{X,Y,Z\}}(t) = \begin{cases}t^{18} & 0 \leq t \leq 1 \\1 & t > 1 \\0 & t < 0\end{cases}$$To find the PDF, we differentiate the CDF:$$f_{\max\{X,Y,Z\}}(t) = \begin{cases}18t^{17} & 0 \leq t \leq 1 \\0 & \text{otherwise}\end{cases}$$which is equivalent to $
  • #1
WMDhamnekar
MHB
376
28
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?
 
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  • #2
Dhamnekar Winod said:
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.
 
  • #3
steep said:
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.
 
  • #4
Dhamnekar Winod said:
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
 
  • #5
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?
 
  • #6
Dhamnekar Winod said:
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 $t \lt 0$
 
  • #7
steep said:
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 $t \lt 0$
 

Related to Distribution and Density functions of maximum of random variables

1. What is a distribution function for maximum of random variables?

A distribution function for maximum of random variables is a mathematical function that describes the probability of a maximum value occurring in a set of random variables. It is typically used to model the likelihood of extreme events or outcomes.

2. How is a density function for maximum of random variables different from a distribution function?

A density function for maximum of random variables is a function that describes the probability density of a maximum value occurring in a set of random variables. It is similar to a distribution function, but it measures the probability of a maximum value at a specific point, rather than over a range of values.

3. What are some common applications of distribution and density functions for maximum of random variables?

Distribution and density functions for maximum of random variables are commonly used in fields such as finance, engineering, and environmental science. They can be used to model extreme events, such as stock market crashes, natural disasters, or equipment failures, and to make predictions about their likelihood.

4. How are distribution and density functions for maximum of random variables calculated?

The calculation of distribution and density functions for maximum of random variables depends on the specific variables and their underlying distributions. In general, they can be calculated using mathematical formulas or statistical software, based on the properties of the random variables and their relationships.

5. What is the significance of understanding distribution and density functions for maximum of random variables?

Understanding distribution and density functions for maximum of random variables is important for accurately modeling and predicting extreme events. By analyzing the probability of maximum values occurring in a set of random variables, scientists can make informed decisions and mitigate potential risks in various fields.

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