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.


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 $t=s=1$.

  • $\begingroup$ 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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