MHB Transformation of a random variable (exponential)

[Jameson]

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]

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

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

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

 

[I like Serena]

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]

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 $$-\frac{\ln(1-y)}{\lambda}$$ for $y \in [-\infty,1)$.

 

[I like Serena]

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 $$-\frac{\ln(1-y)}{\lambda}$$ 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]

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 $$\lim_{x \rightarrow -\infty}F(x)=0$$ and $$\lim_{x \rightarrow \infty}F(x)=1$$. 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?

 

[I like Serena]

I still say no. All CDFs must be increasing or non-decreasing. They must satisfy the property that $$\lim_{x \rightarrow -\infty}F(x)=0$$ and $$\lim_{x \rightarrow \infty}F(x)=1$$. 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]

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

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

 

[I like Serena]

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

That should be $$1-e^{-\lambda A}$$.

 

[Jameson]

That should be $$1-e^{-\lambda A}$$.

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

 

[I like Serena]

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]

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

We start with $$P[X \le F^{-1}(y)]$$ and end up with $$P[X \le F\left( F^{-1}(y) \right)$$. 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?

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

 

[I like Serena]

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

We start with $$P[X \le F^{-1}(y)]$$ and end up with $$P[X \le F\left( F^{-1}(y) \right)$$. 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]

Yep!

Awesome. Thank you! I didn't feel very good about transformations before but now I feel much more confident.

 

[Jameson]

Something I didn't notice before was that $$P[X \le F^{-1}(y)] = F\left( F^{-1}(y) \right)$$ 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 $0 < y <1$.