Applied Stochastic Processess?

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The discussion focuses on the properties of a random variable X that follows a Gaussian distribution, specifically its moments and transformations. The first three moments calculated are E(X) = μ, E(X^2) = σ^2 + μ^2, and E(X^3) = 0, confirming that odd moments are zero when μ ≠ 0. The transformations Y = X + b and Z = σX are also Gaussian, with Y having mean mb and Z having mean σμ. The distribution of W = X^2 follows a Chi-squared distribution, which requires further clarification from participants.

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  • Understanding of Gaussian distribution and its properties
  • Knowledge of statistical moments and their calculations
  • Familiarity with linear transformations of random variables
  • Basic concepts of Chi-squared distribution
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  • Review the properties of Gaussian distributions in detail
  • Study the calculation of statistical moments for various distributions
  • Learn about linear transformations of random variables and their implications
  • Explore the Chi-squared distribution and its applications in statistics
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Random variable X is distributed according to Gaussian distribution
P(x)= (1/sqrt(2πσ^2) * exp(- (x-μ)^2 / (2σ^2) )
1. Calculate its first three moments
2. What are the distributions of Y= X+b; Z=σX; W=X^2


P(x)= (1/sqrt(2πσ^2) * exp(- (x-μ)^2 / (2σ^2) )
so
1.1.
the moment of first degree is:
E(X) = Int (on |R of) x dP(x) = m

1.2. moment of second degree:
we have Var(X) = σ^2 = E(X^2) - E(X)^2 = E(X^2) - m^2
so the moment of degree 2 is
E(X^2) = σ^2 m^2

1.3.moment of third order :
as P is an even function :
E(X^3) = Int ( on |R of) x^3 P(x) dx = 0
and it 's the case for any odd moment n : n = 2k 1

2.
the gaussian characteristic is stable through linear transformation :
Y = X b is gaussian
ith the same σ
and with an expectence of E(Y) = m b
so Y is a N(m b ; σ)

Z = σX is gaussian too :
E(Z) = σm and Var(Z) = σ^2 Var(X) = σ^4
Z is therefore a N(σm , σ^2) (σ^2 is its standard deviation

W = X^2 follws a Khi-squared distribution law ;


CAN SOMEONE CHECK MY METHODS AND HELP ME TO DO LAST QUESTION W=X^2. THANK YOU
 
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ra_forever8 said:
Random variable X is distributed according to Gaussian distribution
P(x)= (1/sqrt(2πσ^2) * exp(- (x-μ)^2 / (2σ^2) )
1. Calculate its first three moments
2. What are the distributions of Y= X+b; Z=σX; W=X^2


P(x)= (1/sqrt(2πσ^2) * exp(- (x-μ)^2 / (2σ^2) )
so
1.1.
the moment of first degree is:
E(X) = Int (on |R of) x dP(x) = m

1.2. moment of second degree:
we have Var(X) = σ^2 = E(X^2) - E(X)^2 = E(X^2) - m^2
so the moment of degree 2 is
E(X^2) = σ^2 m^2

1.3.moment of third order :
as P is an even function :
E(X^3) = Int ( on |R of) x^3 P(x) dx = 0
and it 's the case for any odd moment n : n = 2k 1

2.
the gaussian characteristic is stable through linear transformation :
Y = X b is gaussian
ith the same σ
and with an expectence of E(Y) = m b
so Y is a N(m b ; σ)

Z = σX is gaussian too :
E(Z) = σm and Var(Z) = σ^2 Var(X) = σ^4
Z is therefore a N(σm , σ^2) (σ^2 is its standard deviation

W = X^2 follws a Khi-squared distribution law ;


CAN SOMEONE CHECK MY METHODS AND HELP ME TO DO LAST QUESTION W=X^2. THANK YOU

1.1. Determine the value of m, in terms of μ and σ.
1.2. From σ^2 = E(X^2) - (E X)^2 we do NOT get E(X^2) = σ^2 * (E X)^2, which is what you wrote. Did you mean that, or was it a typo?
1.3. P(x) is most definitely not an even function if μ ≠ 0. Start over.

I don't know what you are doing in 2. You were asked about X+b but instead, you told us about Xb. Are you sure you were asked about σX? A more meaningful question would be about (1/σ)X. As for X^2: just apply the standard formulas you find in your textbook or course notes. It might be quite complicated; much easier would be the case of X^2 if we had μ = 0.
 

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