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elixir [45]
3 years ago
9

Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.

Mathematics
1 answer:
Cloud [144]3 years ago
3 0

If F_Y(y) is the cumulative distribution function for Y, then

F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)

Then the probability density function for Y is f_Y(y)={F_Y}'(y):

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}

The nth moment of Y 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

Let u=\ln y, so that \mathrm du=\frac{\mathrm dy}y and y^n=e^{nu}:

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

Complete the square in the exponent:

nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2

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

But \frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2} is exactly the PDF of a normal distribution with mean n and variance 1; in other words, the 0th moment of a random variable U\sim N(n,1):

E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1

so we end up with

E[Y^n]=e^{\frac12n^2}

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Step-by-step explanation:

call the number x, and the other number y, the answer is:

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2 years ago
Which could be the resulting equation when elimination is used to solve the given system of equations ?
WINSTONCH [101]

Answer:

10·b = 60

Step-by-step explanation:

The given system of equation is presented as follows;

5·a + 5·b = 25...(1)

-5·a + 5·b = 35...(2)

Given that the coefficient of a in equation (1) is equal in magnitude but opposite in sign to the coefficient of 'a' in equation (2), to eliminate the variable 'a' when using the elimination method, we add both equations as follows;

5·a + 5·b + (-5·a + 5·b) = 25 + 35 = 60

5·a - 5·a + 5·b + 5·b = 60

5·a - 5·a = 0

5·b + 5·b = 10·b

∴ 5·a - 5·a + 5·b + 5·b = 60 = 0 + 10·b = 10·b

∴ 10·b = 60

5 0
3 years ago
-1≤ 3x-10≤ 2 please help me
kicyunya [14]

Answer:

3 ≤ x ≤ 4

Step-by-step explanation:

Step 1: Add 10 to all parts/sections.

  • -1 + 10 \leq  3x - 10 + 10 \leq 2+10
  • 9\leq 3x\leq 12

Step 2: Divide all parts/sections by 3.

  • \frac{9}{3} \leq \frac{3x}{3} \leq \frac{12}{3}
  • 3 \leq x \leq  4

Therefore, the answer is 3 ≤ x ≤ 4.

5 0
2 years ago
Read 2 more answers
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