# [SOLVED]Transformation of a random variable (exponential)

#### Jameson

Staff member
Problem: Suppose that $X \text{ ~ Exp}(\lambda)$ and denote its distribution function by $F$. What is the distribution of $Y=F(X)$?

My attempt: First off, I'm assuming this is asking for the CDF of $Y$. Sometimes it's not clear what terminology refers to the PDF or the CDF for me.

$P[Y \le y]= P[F(X) \le y]= P[X \le F^{-1}(y)]$

For $x \ge 0$ the CDF of the exponential distribution is $1-e^{-\lambda x}$. So do I need to find the inverse of this and go from there? I have a feeling this is the right direction or this problem is trivially simple.

#### Jameson

Staff member
I've found the inverse of $F(X)$. It is $$\displaystyle F^{-1}(X)=-\frac{\ln(1-x)}{\lambda}$$

From my OP, we have: $$\displaystyle P[Y \le y]= P[F(X) \le y]= P[X \le F^{-1}(y)]=P \left[X \le -\frac{\ln(1-y)}{\lambda} \right]$$

This inverse is only defined though for $y<1$. Is that all? I feel like I'm missing something.

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
It's ambiguous for me as well.
The use of capitals suggests the cumulative distribution function is intended.
It would be nice if they made that explicit.

Anyway, I think it is a bit of a trick question, since the function is decreasing.
That has some impact on the inequalities...

And you didn't give a final formula.

#### Jameson

Staff member
It's ambiguous for me as well.
The use of capitals suggests the cumulative distribution function is intended.
It would be nice if they made that explicit.

Anyway, I think it is a bit of a trick question, since the function is decreasing.
That has some impact on the inequalities...

And you didn't give a final formula.
The density of an exponential random variable is decreasing but the CDF isn't. The CDF is our $Y$.

What final formula? I think it should be $$\displaystyle -\frac{\ln(1-y)}{\lambda}$$ for $y \in [-\infty,1)$.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
The density of an exponential random variable is decreasing but the CDF isn't. The CDF is our $Y$.
The CDF is $F(x)=1-e^{-\lambda x}$.
Isn't that a decreasing function?

What final formula? I think it should be $$\displaystyle -\frac{\ln(1-y)}{\lambda}$$ for $y \in [-\infty,1)$.
You ended with $F_Y(y) = P(X \le \text{some expression})$.
But $P(X \le x) = 1-e^{-\lambda x}$.
Doesn't that mean there is some more work to do?

#### Jameson

Staff member
The CDF is $F(x)=1-e^{-\lambda x}$.
Isn't that a decreasing function?
I still say no. All CDFs must be increasing or non-decreasing. They must satisfy the property that $$\displaystyle \lim_{x \rightarrow -\infty}F(x)=0$$ and $$\displaystyle \lim_{x \rightarrow \infty}F(x)=1$$. Plus if you look at a graph of it you can see that it is increasing and has a ceiling of 1.

You ended with $F_Y(y) = P(X \le \text{some expression})$.
But $P(X \le x) = 1-e^{-\lambda x}$.
Doesn't that mean there is some more work to do?
Hmm, I thought of it like:

$$P[Y \le y]=P \left[ X \le \displaystyle -\frac{\ln(1-y)}{\lambda}\right]$$

Where should I head from here?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I still say no. All CDFs must be increasing or non-decreasing. They must satisfy the property that $$\displaystyle \lim_{x \rightarrow -\infty}F(x)=0$$ and $$\displaystyle \lim_{x \rightarrow \infty}F(x)=1$$. Plus if you look at a graph of it you can see that it is increasing and has a ceiling of 1.
My mistake. You are right.

Hmm, I thought of it like:

$$P[Y \le y]=P \left[ X \le \displaystyle -\frac{\ln(1-y)}{\lambda}\right]$$

Where should I head from here?
$$P[ X \le \text{some expression}] = 1 - e^{-\lambda \cdot (\text{some expression})}$$

#### Jameson

Staff member
$$\displaystyle P[X \le F^{-1}(y)]=1-e^{-Ax}$$, where $$\displaystyle A=-\frac{\ln(1-y)}{\lambda}$$?

EDIT: No, that isn't right. I think I'm just supposed to evaluate $$\displaystyle F^{-1}(y)$$.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
$$\displaystyle P[X \le F^{-1}(y)]=1-e^{-Ax}$$, where $$\displaystyle A=-\frac{\ln(1-y)}{\lambda}$$?
That should be $$\displaystyle 1-e^{-\lambda A}$$.

#### Jameson

Staff member
That should be $$\displaystyle 1-e^{-\lambda A}$$.
Agreed, however I still don't see how that is equal to $F^{-1}(y)$. The above seems to be $F\left( F^{-1}(y) \right)$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Agreed, however I still don't see how that is equal to $F^{-1}(y)$.
It isn't.

$A=F^{-1}(y)$

The above seems to be $F\left( F^{-1}(y) \right)$
Neat! Isn't it?

#### Jameson

Staff member
I just don't see how you get from $$\displaystyle F^{-1}(y)$$ to $$\displaystyle F\left( F^{-1}(y) \right)$$.

We start with $$\displaystyle P[X \le F^{-1}(y)]$$ and end up with $$\displaystyle P[X \le F\left( F^{-1}(y) \right)$$. Can you explain how we can make that jump?

Wait! I think I might have answered my own question. The definition of $F(X)=P[X \le x]$ so we just use that definition correct?

$$P[X \le F^{-1}(y)] = F\left( F^{-1}(y) \right)$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I just don't see how you get from $$\displaystyle F^{-1}(y)$$ to $$\displaystyle F\left( F^{-1}(y) \right)$$.

We start with $$\displaystyle P[X \le F^{-1}(y)]$$ and end up with $$\displaystyle P[X \le F\left( F^{-1}(y) \right)$$. Can you explain how we can make that jump?

Wait! I think I might have answered my own question. The definition of $F(X)=P[X \le x]$ so we just use that definition correct?

$$P[X \le F^{-1}(y)] = F\left( F^{-1}(y) \right)$$
Yep!

#### Jameson

Something I didn't notice before was that $$\displaystyle P[X \le F^{-1}(y)] = F\left( F^{-1}(y) \right)$$ simplifies to $y$. Does that mean that the distribution of $Y$ is $y$?
EDIT: I think the only other thing is to describe the bounds. For the exponential distribution the CDF is non-zero for $x \ge 0$ and 0 for $x<0$. I think that means in terms of $Y$ the CDF is $y$, where $0 < y <1$.