Probability of exponential of normal distribution

Question

Problem : Let $Y$ is normal random variable with mean of $10$ and a variance of $100$.

If $X=e^Y$, Compute $P(e^{10}<X<e^{20})$.

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My Attempt:

Since $X=e^Y$, rewrite probability : $P(e^{10}<e^Y<e^{20})$.

Since a map $x \to e^x$ is bijective, $$P(e^{10}<e^Y<e^{20}) = P(10<Y<20)$$

Then substitute $Z=\frac{Y-10}{10}$, then $Z$ is standard normal random variable and $$P(10<Y<20) = P(0<Z<1)$$

Finally with normal table, $$P(0<Z<1) = \Phi(1)-\Phi(0) \approx .3413 $$ where $\Phi$ is CDF of standard normal distribution.

Is my attempt correct? If not, what should I modify?

Thanks for help.

Answer 1

Looks good to me! However, just because a function is bijective doesn't mean you can take that step. It should be monotone increasing/decreasing.

Consider this bijective function: $f(1) = 0, f(2) = 10, f(3) = 9, f(4) = 8$

$f(1) < y < f(3)$ does NOT mean
$f^{-1}(f(1)) < f^{-1}(y) < f^{-1}(f(3))$
$1 < f^{-1}(y) < 3$
because we could have taken $y=8, f^{-1}(y)=4$, for example.