My Math Forum (normal distribution) E(e^x) and median of e^x of N(2,5)

 February 8th, 2018, 02:11 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, 03:53 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 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, 04:28 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, 11:47 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, 03:47 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 $\tau = e^\mu = e^2$

With respect to finding E(e^x) --> I used the following formula:
e^(u+0,5*variance) which solved the problem.

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