False 2(18)-3/3= 35 and 2/3 = 0.6 Repeating
If
is the cumulative distribution function for
, then
![F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)](https://tex.z-dn.net/?f=F_Y%28y%29%3DP%28Y%5Cle%20y%29%3DP%28e%5EX%5Cle%20y%29%3DP%28X%5Cle%5Cln%20y%29%3DF_X%28%5Cln%20y%29)
Then the probability density function for
is
:
![f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}](https://tex.z-dn.net/?f=f_Y%28y%29%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dy%7DF_X%28%5Cln%20y%29%3D%5Cdfrac1yf_X%28%5Cln%20y%29%3D%5Cbegin%7Bcases%7D%5Cfrac1%7By%5Csqrt%7B2%5Cpi%7D%7De%5E%7B-%5Cfrac12%28%5Cln%20y%29%5E2%7D%26%5Ctext%7Bfor%20%7Dy%3E0%5C%5C0%26%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
The
th moment of
is
![E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20y%5Enf_Y%28y%29%5C%2C%5Cmathrm%20dy%3D%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_0%5E%5Cinfty%20y%5E%7Bn-1%7De%5E%7B-%5Cfrac12%28%5Cln%20y%29%5E2%7D%5C%2C%5Cmathrm%20dy)
Let
, so that
and
:
![E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7Bnu%7De%5E%7B-%5Cfrac12u%5E2%7D%5C%2C%5Cmathrm%20du%3D%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7Bnu-%5Cfrac12u%5E2%7D%5C%2C%5Cmathrm%20du)
Complete the square in the exponent:
![nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2](https://tex.z-dn.net/?f=nu-%5Cdfrac12u%5E2%3D-%5Cdfrac12%28u%5E2-2nu%2Bn%5E2-n%5E2%29%3D%5Cdfrac12n%5E2-%5Cdfrac12%28u-n%29%5E2)
![E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B%5Cfrac12%28n%5E2-%28u-n%29%5E2%29%7D%5C%2C%5Cmathrm%20du%3D%5Cfrac%7Be%5E%7B%5Cfrac12n%5E2%7D%7D%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cfrac12%28u-n%29%5E2%7D%5C%2C%5Cmathrm%20du)
But
is exactly the PDF of a normal distribution with mean
and variance 1; in other words, the 0th moment of a random variable
:
![E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1](https://tex.z-dn.net/?f=E%5BU%5E0%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cfrac12%28u-n%29%5E2%7D%5C%2C%5Cmathrm%20du%3D1)
so we end up with
![E[Y^n]=e^{\frac12n^2}](https://tex.z-dn.net/?f=E%5BY%5En%5D%3De%5E%7B%5Cfrac12n%5E2%7D)
Step-by-step explanation: Okay so ther is 2 blue and 1 red and 5 white. 1/8 chance you would get red and then 2nd try 2/8 chance so... Im pretty sure its 1/32
Hey There :)
This is because one is half of two. You could also multiply 2 * 0.5= 1, 0.5 represents
![\frac{1}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20)
and two is the whole number. This is how
![\frac{1}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20)
of two equals one.
Have A Brainly Day :)