# Probability density function of uniform distribution

• diracdelta
In summary, the probability density function of a random variable Y = \sqrt{X} + 1 is given by: F_{Y}(y)=P(Y\leq y)=P(x^{2}+1\leq y)=P(x\leq y^{2}+1)=F_{x}(y^{2}+1) \\ f_{X}(x)=F_{X}'(y^{2}+1)=f_{X}(y^{2}+1)* \left | \frac{d}{dy} (y^{2}+1)\right |=f_{X}(y^{2}+1)*2
diracdelta

## Homework Statement

Random variable X is uniformly distributed on interval [0,1]:
$$f(x)=\begin{cases} 1 & \text{ if } 0\leq x\leq 1\\ 0 & \text{ else} \end{cases}$$

a) Find probability density function ρ(y) of random variable $$Y=\sqrt{X} +1$$

I tried like this. Is it good, if no why not?

$$F_{Y}(y)=P(Y\leq y)=P(x^{2}+1\leq y)=P(x\leq y^{2}+1)=F_{x}(y^{2}+1) \\ f_{X}(x)=F_{X}'(y^{2}+1)=f_{X}(y^{2}+1)* \left | \frac{d}{dy} (y^{2}+1)\right |=f_{X}(y^{2}+1)*2y \\ f_{Y}(y)=\begin{cases} 2y & \text{ if } 1\leq x\leq 2\\ 0 & \text{ else} \end{cases}$$

diracdelta said:

## Homework Statement

Random variable X is uniformly distributed on interval [0,1]:
$$f(x)=\begin{cases} 1 & \text{ if } 0\leq x\leq 1\\ 0 & \text{ else} \end{cases}$$

a) Find probability density function ρ(y) of random variable $$Y=\sqrt{X} +1$$

I tried like this. Is it good, if no why not?

$$F_{Y}(y)=P(Y\leq y)=P(x^{2}+1\leq y)=P(x\leq y^{2}+1)=F_{x}(y^{2}+1) \\ f_{X}(x)=F_{X}'(y^{2}+1)=f_{X}(y^{2}+1)* \left | \frac{d}{dy} (y^{2}+1)\right |=f_{X}(y^{2}+1)*2y \\ f_{Y}(y)=\begin{cases} 2y & \text{ if } 1\leq x\leq 2\\ 0 & \text{ else} \end{cases}$$
How did $\sqrt{X}+ 1$ become $X^2+ 1$? If you intended the inverse function then it should be $X= (Y- 1)^2$

diracdelta
Hmm, i see.
Ok then, i get this
$$F_{Y}(y)=P(Y\leq y)=P(x^{1/2}+1\leq y)=P(x\leq (y+1)^{2}) =F_{x}(y+1)^{2} \\ f_{X}(x)=F_{X}'(y+1)^{2}=f_{X}(y+1)^{2}* \left | \frac{d}{dy} (y+1)^{2})\right |=f_{X}(y+1)^{2})*2y \\ f_{Y}(y)=\begin{cases} 2y & \text{ if } 1\leq x\leq 2\\ 0 & \text{ else} \end{cases}$$[/QUOTE]

But what about y's probability density function ?
Is it ok?

diracdelta said:
##F_{Y}(y)=P(Y\leq y)=P(x^{\frac 12}+1\leq y)##
That should really be
##F_{Y}(y)=P(Y\leq y)=P(X^{\frac 12}+1\leq y)##
diracdelta said:
##=P(x\leq (y+1)^{2}) ##
Check that step.
diracdelta said:
##f_{X}(x)=F_{X}'(y+1)^{2}=f_{X}(y+1)^{2}* \left | \frac{d}{dy} (y+1)^{2})\right |##
##F_{X}'((y+1)^{2})## just means the derivative wrt x of FX(x) evaluated at x = (y+1)2
##f_{X}(x)=F_{X}'((y+1)^{2})=f_{X}((y+1)^{2})##

diracdelta said:
Hmm, i see.
Ok then, i get this
$$F_{Y}(y)=P(Y\leq y)=P(x^{1/2}+1\leq y)=P(x\leq (y+1)^{2}) =F_{x}(y+1)^{2} \\ f_{X}(x)=F_{X}'(y+1)^{2}=f_{X}(y+1)^{2}* \left | \frac{d}{dy} (y+1)^{2})\right |=f_{X}(y+1)^{2})*2y \\ f_{Y}(y)=\begin{cases} 2y & \text{ if } 1\leq x\leq 2\\ 0 & \text{ else} \end{cases}$$

But what about y's probability density function ?
Is it ok?[/QUOTE]

How do you get from ##x^{1/2}+1 \leq y## to ##x \leq (y+1)^2##? (I don't.)

Anyway, since ##X## ranges from 0 to 1, so does ##\sqrt{X}## and that means that ##Y = 1 + \sqrt{X}## ranges from 1 to 2. So, you ought to have ##F_Y(1) = 0## and ##F_Y(2) = 1##. Do you have that?

Last edited:
diracdelta said:
$$F_{Y}(y)=P(Y\leq y)=P(x^{1/2}+1\leq y)=P(x\leq (y+1)^{2}) =F_{x}(y+1)^{2}$$
Up to here it's OK, except for the two errors pointed out by haruspex. Fix those and you will have a correct formula for ##F_Y(y)## in terms of ##F_X## applied to some function of ##y##.

Next you can use the fact that ##F_X(x)=\int_0^x f_X(u)du=\int_0^x du=x## to write a formula for ##F_{Y}(y)## in terms of ##y##.

Forget about everything you wrote below your first line, as that heads off on a goose chase involving ##f_X## and there is no need to refer to ##f_X## any further.

Differentiating your formula for ##F_Y(y)## will give you a formula for ##f_Y(y)##.

To complete the answer you need to put range constraints into your expression for ##f_Y##, along the lines suggested by Ray.

## 1. What is a probability density function (PDF)?

A probability density function is a mathematical function that describes the likelihood of a continuous random variable taking on a specific value within a given range. It is used to model the probability distribution of a continuous random variable.

## 2. How is a uniform distribution defined?

A uniform distribution is a probability distribution where all values within a given range have an equal probability of occurring. This means that the probability density function of a uniform distribution is constant for all values within the range.

## 3. How is the probability density function of a uniform distribution calculated?

The probability density function of a uniform distribution is calculated by dividing 1 by the range of values. This ensures that the total area under the curve is equal to 1, representing the total probability of all possible outcomes.

## 4. What is the difference between a continuous and discrete uniform distribution?

A continuous uniform distribution is used to model continuous random variables, meaning that the variable can take on any value within a given range. A discrete uniform distribution, on the other hand, is used for discrete random variables, which can only take on a finite set of values within a given range.

## 5. How is the probability density function of a uniform distribution used in real-world applications?

The probability density function of a uniform distribution is commonly used in statistics and probability theory to model a wide range of phenomena, such as stock prices, weather patterns, and population growth. It is also used in fields such as physics, engineering, and economics to analyze and predict various outcomes and events.

Replies
4
Views
757
Replies
2
Views
1K
Replies
7
Views
959
Replies
9
Views
739
Replies
5
Views
1K
Replies
6
Views
1K
Replies
7
Views
1K
Replies
7
Views
1K
Replies
14
Views
1K
Replies
15
Views
1K