 My Math Forum (normal distribution) E(e^x) and median of e^x of N(2,5)
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 February 8th, 2018, 01:11 AM #1 Newbie   Joined: Sep 2017 From: xxxx Posts: 8 Thanks: 0 (normal distribution) E(e^x) and median of e^x of N(2,5) Dear all, Started with a statistics course and even the basic concepts are a problem for me. Reason for that is that there are no examples in the book and just a bunch of formulas. It makes it hard to grasp the concept. I hope one can help me out! So the question is: IF x ~ N(2,5) what is E(e^x) and what is the median of e^x? Where is x is normally distributed. According to the solutions, the answer is 90.017 and 7.3891 respectively but there is no explanation! Can someone explain to me how I should approach this is plain English? any help is appreciated Kind regards  February 8th, 2018, 02:53 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 if $X$ is distributed as $f_X(x)$ then $E[g(x)] = \displaystyle \int \limits_{-\infty}^\infty~g(x) f_X(x)~dx$ here $f_X(x) = \dfrac{1}{5\sqrt{2\pi}}e^{-\left(\dfrac{(x-2)^2}{25}\right)}$ $g(x) = e^x$ I leave the integration to you. When $f_X(x)$ is continuous the median value of $g(x)$ occurs at $\tau$ where $P[g(x) < \tau] = P[g(x) > \tau]$ This leads to the integral equation $\displaystyle \int_{-\infty}^\tau ~g(x)f_X(x)~dx = \int_\tau^\infty ~g(x)f_X(x)~dx$ which can be numerically solved for $\tau$ Again I leave this for you. February 8th, 2018, 03:28 AM   #3
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Quote:
 Originally Posted by romsek When $f_X(x)$ is continuous the median value of $g(x)$ occurs at $\tau$ where $P[g(x) < \tau] = P[g(x) > \tau]$ This leads to the integral equation $\displaystyle \int_{-\infty}^\tau ~g(x)f_X(x)~dx = \int_\tau^\infty ~g(x)f_X(x)~dx$ which can be numerically solved for $\tau$ Again I leave this for you.
No, need for anything numerical:
If $\mathbb{P}[g(X)<\tau] = 1/2$, then $\mathbb{P}[X<g^{-1}(\tau)] = 1/2$. But then $g^{-1}(\tau)$ is the median of $X$. February 8th, 2018, 10:47 AM   #4
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Quote:
 Originally Posted by romsek if $X$ is distributed as $f_X(x)$ then $E[g(x)] = \displaystyle \int \limits_{-\infty}^\infty~g(x) f_X(x)~dx$ here $f_X(x) = \dfrac{1}{5\sqrt{2\pi}}e^{-\left(\dfrac{(x-2)^2}{25}\right)}$ $g(x) = e^x$ I leave the integration to you. When $f_X(x)$ is continuous the median value of $g(x)$ occurs at $\tau$ where $P[g(x) < \tau] = P[g(x) > \tau]$ This leads to the integral equation $\displaystyle \int_{-\infty}^\tau ~g(x)f_X(x)~dx = \int_\tau^\infty ~g(x)f_X(x)~dx$ which can be numerically solved for $\tau$ Again I leave this for you.
The calculation for the median above is incorrect. I apologize.

What I should have done was let

$Y = e^X$

a quick calculation which you should be able to replicate shows

$f_Y(y) = \dfrac{f_X(\ln(y))}{y}$

This is the log-normal distribution.

Now as both $f_X(x)$ and $f_Y(y)$ are continuous we can find the median, $\tau$ of $Y$ as

$\displaystyle \int \limits_0^\tau ~f_Y(y)~dy = \dfrac 1 2$

This can be solved to obtain

$\tau = e^\mu = e^2$ February 10th, 2018, 02:47 AM   #5
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Quote:
 Originally Posted by romsek The calculation for the median above is incorrect. I apologize. What I should have done was let $Y = e^X$ a quick calculation which you should be able to replicate shows $f_Y(y) = \dfrac{f_X(\ln(y))}{y}$ This is the log-normal distribution. Now as both $f_X(x)$ and $f_Y(y)$ are continuous we can find the median, $\tau$ of $Y$ as $\displaystyle \int \limits_0^\tau ~f_Y(y)~dy = \dfrac 1 2$ This can be solved to obtain $\tau = e^\mu = e^2$

Thank you for the answer! Your last answer helped me a lot.
With respect to finding E(e^x) --> I used the following formula:
e^(u+0,5*variance) which solved the problem.  Tags distribution, eex, median, normal Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mathmathy Probability and Statistics 6 August 16th, 2017 05:29 AM faker97 Advanced Statistics 1 May 2nd, 2017 10:51 AM froydipj Probability and Statistics 3 February 29th, 2016 04:35 PM nakys Advanced Statistics 0 October 3rd, 2013 08:27 AM koharudin Advanced Statistics 0 November 18th, 2009 05:43 PM

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