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The problem seems simple:

Let X be an exponential random variable such that $P(X \le 2) = 2P(X > 4)$. Find the variance of X.

Easy, right? $ P(x \le 2) = 1 - e^{-2\lambda} $ and $ P(x > 4) = e^{-4\lambda} $

Assuming that is right, which, it's possible that it is not...

I can't find any way to solve $$ 1 - e^{-2\lambda} = 2e^{-4\lambda} $$ for $\lambda$.

I am rusty on my Exponential and logarithmic function solving, but it seems unsolvable.

I have tried taking $ln$ of both sides. Then what do you do with $ln(1 - e^{-2\lambda})$?

I have tried rewriting 1 as $e^{0}$ but that didn't get me any further.

I know the variance is $\frac{4}{(ln 2) ^2}$ which means $\lambda = \frac {ln 2} {2}$. And that checks out with the equation above ($.5 = .5$). But how?

Thank you!

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Make the substitution $u = e^{-2\lambda}$, and solve the resulting quadratic equation. Once you have values for $u$, convert them back in to $\lambda$. Hint: there's only one solution. –  Sammy Black Mar 17 '13 at 7:02

1 Answer 1

up vote 2 down vote accepted

It looks like a quadratic equation with respect to $e^{-2\lambda}$. The only positive solution of $1-x=2x^2$ is $\frac12$, so $e^{-2\lambda}=\frac12$, which gives the desired answer.

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Wow! I don't think I have ever solved a quadratic with respect to a variable in an exponent position. Another trick in the arsenal. TYVM. –  Katarzyna Mar 17 '13 at 7:08
    
You're welcome. –  mne__povezlo Mar 17 '13 at 7:13

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