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The exercise problem that I refer to is from the textbook "Introduction to Probability (2e) - Blitzstein & Hwang."

I was studying probability when I came across a problem that I believe I solved correctly, but have been getting the incorrect answer for. Here's the specific problem:


Let $Y = X^\beta$, with $X \sim$ Expo($1$) and $\beta \gt 0$. $Y$ is called the Weibull distribution with parameter $\beta$. For this problem, let $\beta = 3$.

Find $P(Y \gt s + t\ |\ Y \gt s)$ for $s,\ t \gt 0$. Does $Y$ have the memoryless property?


My Solution

The CDF of $Y$ is as follows:

\begin{align} P(Y \le x) & = P(X^3 \le x) \\ & = P(X \le x^{\frac{1}{3}}) \\ & = 1 - e^{-x^\frac{1}{3}} \end{align}

Using the CDF, we can solve the given conditional probability.

\begin{align} P(Y \gt s + t\ |\ Y \gt s) & = \frac{P(Y \gt s + t,\ Y \gt s)}{P(Y \gt S)}\\ & = \frac{P(Y \gt s + t)}{P(Y \gt s)} \\ & = \frac{1 - P(Y \le s + t)}{1 - P(Y \le s)} \\ & = \frac{e^{-(s + t)^{\frac{1}{3}}}}{e^{-s^\frac{1}{3}}} \\ & = e^{-t^{\frac{1}{3}}} \\ & = 1 - (1 - e^{-t^{\frac{1}{3}}}) \\ & = 1 - F_Y(t) \\ & = P(Y \gt t) \end{align}

Therefore, I concluded that $Y$ in fact does have the memoryless property.


However, the answer that I've checked states that $Y$ does not have the memoryless property.

Is there something wrong with the solution that I've come up with?


Any feedback is appreciated. Thank you.

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Your calculation is not correct. You are writing $e^{-(s+t)^{1/3}}$ as $e^{s^{1/3}}$ times $e^{-t^{1/3}}$ which is not true. For example this is false for $t=s=1$.

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  • $\begingroup$ Oh, I see the mistake I was making. Thank you for the correction! $\endgroup$ – Seankala Nov 1 '18 at 12:23

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