If the random variable V ∼ N ( μ , σ 2 ) , σ 2 > 0 , V \sim N(\mu, \sigma^2),\sigma^2>0, V ∼ N ( μ , σ 2 ) , σ 2 > 0 , then the random variable W = ( X − μ ) 2 σ 2 = Z 2 ∼ χ 2 ( 1 ) , W=\dfrac{(X-\mu)^2}{\sigma^2}=Z^2\sim \chi^2(1), W = σ 2 ( X − μ ) 2 = Z 2 ∼ χ 2 ( 1 ) , chi-squared distribution, with ν = 1 \nu=1 ν = 1 degrees of freedom.
Suppose X ∼ N ( 0 , 1 ) , X\sim N(0, 1), X ∼ N ( 0 , 1 ) , then Y = ( X − 0 ) 2 1 2 = X 2 ∼ χ 2 ( 1 ) , Y=\dfrac{(X-0)^2}{1^2}=X^2\sim \chi^2(1), Y = 1 2 ( X − 0 ) 2 = X 2 ∼ χ 2 ( 1 ) , chi-squared distribution, with ν = 1 \nu=1 ν = 1 degrees of freedom.
Its density function is given by
f ( y ; ν ) = { 1 2 ν / 2 Γ ( ν / 2 ) y ν / 2 − 1 e − y / 2 , y > 0 0 , elsewhere f(y;\nu)= \begin{cases}
\dfrac{1}{2^{\nu/2}\Gamma(\nu/2)}y^{\nu/2-1}e^{-y/2},y>0 \\
0, \text{elsewhere}
\end{cases} f ( y ; ν ) = ⎩ ⎨ ⎧ 2 ν /2 Γ ( ν /2 ) 1 y ν /2 − 1 e − y /2 , y > 0 0 , elsewhere Then
f ( y ; ν ) = { 1 2 π y e − y / 2 , y > 0 0 , elsewhere f(y;\nu)= \begin{cases}
\dfrac{1}{\sqrt{2\pi}\sqrt{y}}e^{-y/2},y>0 \\
0, \text{elsewhere}
\end{cases} f ( y ; ν ) = ⎩ ⎨ ⎧ 2 π y 1 e − y /2 , y > 0 0 , elsewhere The mean of the distribution is equal to the number of degrees of freedom:
E ( Y ) = μ Y = ν = 1 E(Y)=\mu_Y=\nu=1 E ( Y ) = μ Y = ν = 1 The variance is equal to two times the number of degrees of freedom:
V a r ( Y ) = σ Y 2 = 2 ν = 2 Var(Y)=\sigma_Y^2=2\nu=2 Va r ( Y ) = σ Y 2 = 2 ν = 2
Comments