- #1
Jameson
Gold Member
MHB
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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.
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.