# University of California, Los Angeles Department of Statistics. Distributions related to the normal distribution

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1 Uiversity of Califoria, Los Ageles Departmet of Statistics Statistics 100B Istructor: Nicolas Christou Three importat distributios: Distributios related to the ormal distributio Chi-square (χ ) distributio. t distributio. F distributio. Before we discuss the χ, t, ad F distributios here are few importat thigs about the gamma (Γ) distributio. The gamma distributio is useful i modelig skewed distributios for variables that are ot egative. A radom variable X is said to have a gamma distributio with parameters α, β if its probability desity fuctio is give by f(x) = xα 1 e x β, α, β > 0, x 0. β α Γ(α) E(X) = αβ ad σ = αβ. A brief ote o the gamma fuctio: The quatity Γ(α) is kow as the gamma fuctio ad it is equal to: Γ(α) = Useful result: 0 Γ( 1 ) = π. x α 1 e x dx. If we set α = 1 ad β = 1 λ we get f(x) = λe λx. We see that the expoetial distributio is a special case of the gamma distributio. 1

2 The gamma desity for α = 1,, 3, 4 ad β = 1. Gamma distributio desity f(x) Γ(α = 1, β = 1) Γ(α =, β = 1) Γ(α = 3, β = 1) Γ(α = 4, β = 1) x Momet geeratig fuctio of the X Γ(α, β) radom variable: Proof: M X (t) = (1 βt) α M X (t) = Ee tx = Let y = x( 1 βt ) x = β y, ad dx = β β 1 βt M X (t) = M X (t) = 1 β α Γ(α) 0 0 e tx xα 1 e x β β α Γ(α) dx = 1 x α 1 e β α Γ(α) 0 1 βt x( β ) dx dy. Substitute these i the expressio above: 1 βt ( ) α 1 β y α 1 e y β 1 βt 1 βt dy ( ) α 1 1 β β y α 1 e y dy M β α X (t) = (1 βt) α. Γ(α) 1 βt 1 βt 0

3 Theorem: Let Z N(0, 1). The, if X = Z, we say that X follows the chi-square distributio with 1 degree of freedom. We write, X χ 1. Proof: Fid the distributio of X = Z, where f(z) = 1 π e 1 z. Begi with the cdf of X: F X (x) = P (X x) = P (Z x) = P ( x Z x) F X (x) = F Z ( x) F Z ( x). Therefore: f X (x) = 1 x 1 1 π e 1 x + 1 x 1 1 e 1 x = 1 π 1 π x 1 e x, or f X (x) = x 1 e x 1 Γ( 1 ). This is the pdf of Γ( 1, ), ad it is called the chi-square distributio with 1 degree of freedom. We write, X χ 1. The momet geeratig fuctio of X χ 1 is M X (t) = (1 t) 1. Theorem: Let Z 1, Z,..., Z be idepedet radom variables with Z i N(0, 1). If Y = zi Y follows the chi-square distributio with degrees of freedom. We write Y χ. the Proof: Fid the momet geeratig fuctio of Y. Sice Z 1, Z,..., Z are idepedet, M Y (t) = M Z 1 (t) M Z (t)... M Z (t) Each Z i follows χ 1 ad therefore it has mgf equal to (1 t) 1. Coclusio: M Y (t) = (1 t). This is the mgf of Γ(, ), ad it is called the chi-square distributio with degrees of freedom. Theorem: Let X 1, X,..., X idepedet radom variables with X i N(µ, σ). It follows directly form the previous theorem that if Y = ( ) xi µ the Y χ σ. 3

4 We kow that the mea of Γ(α, β) is E(X) = αβ ad its variace var(x) = αβ. Therefore, if X χ it follows that: E(X) =, ad var(x) =. Theorem: Let X χ ad Y χ m. If X, Y are idepedet the X + Y χ +m. Proof: Use momet geeratig fuctios. Shape of the chi-square distributio: I geeral it is skewed to the right but as the degrees of freedom icrease it becomes N(, ). Here is the graph: Χ 3 f(x) x Χ 10 f(x) x Χ 30 f(x) x 4

5 Example 1 If X χ 16, fid the followig: a. P (X < 8.85). b. P (X > 34.7). c. P (3.54 < X < 8.85). d. If P (X < b) = 0.10, fid b. e. If P (X < c) = 0.950, fid c. The χ distributio - examples Example If X χ 1, fid costats a ad b such that P (a < X < b) = 0.90 ad P (X < a) = Example 3 If X χ 30, fid the followig: a. P (13.79 < X < 16.79). b. Costats a ad b such that P (a < X < b) = 0.95 ad P (X < a) = c. The mea ad variace of X. Example 4 If the momet-geeratig fuctio of X is M X (t) = (1 t) 60, fid: a. E(X). b. V ar(x). c. P (83.85 < X < ). 5

6 Theorem: Let X 1, X,..., X idepedet radom variables with X i N(µ, σ). Defie the sample variace as Proof: S = 1 1 (x i x). The ( 1)S σ χ 1. Example: Let X 1, X,..., X 16 i.i.d. radom variables from N(50, 10). Fid ( ) a. P 796. < (X i 50) < 630. ( b. P 76.1 < ) (X i X) <

7 The χ 1 (1 degree of freedom) - simulatio A radom sample of size = 100 is selected from the stadard ormal distributio N(0, 1). Here is the sample ad its histogram. [1] [5] [9] [13] [17] [1] [5] [9] [33] [37] [41] [45] [49] [53] [57] [61] [65] [69] [73] [77] [81] [85] [89] [93] [97] Histogram of the radom sample of =100 Desity z 7

8 The squared values of the sample above ad their histogram are show below. [1] e e e e+00 [5] e e e e+00 [9].9541e e e e+00 [13] e e e e-01 [17] e e e e-01 [1] e e e e-03 [5] e e e e+00 [9] 5.578e e e e-01 [33] e e e e-01 [37] e e e e+00 [41].33360e e e e-01 [45] e e e e-0 [49] e e e e-01 [53] e e e e-01 [57] e e e e-01 [61] e e e e-01 [65] e e e e-01 [69] e e e e-0 [73] e e e e-01 [77] e e e e+00 [81] e e e e-0 [85].3754e e e e-01 [89] e e e e+00 [93] e e e e-03 [97] e e e e+00 Histogram of the squared values of radom sample of =100 Desity z 8

9 The t distributio Defiitio: Let Z N(0, 1) ad U χ df. If Z, U are idepedet the the ratio Z U df follows the t (or Studet s t) distributio with degrees of freedom equal to df. We write X t df. The probability desity fuctio of the t distributio with df = degrees of freedom is f(x) = +1 Γ( ) ( ) +1 πγ( ) 1 + x, < x <. Let X t. The, E(X) = 0 ad var(x) =. The t distributio is similar to the stadard ormal distributio N(0, 1), but it has heavier tails. However as the t distributio coverges to N(0, 1) (see graph below). f(x) N(0, 1) t 15 t 5 t x 9

10 Applicatio: Let X 1, X,..., X be idepedet ad idetically distributed radom variables each oe havig N(µ, σ). We have see earlier that ( 1)S χ σ 1. We also kow that X µ σ N(0, 1). We ca apply the defiitio of the t distributio (see previous page) to get the followig: X µ σ ( 1)S σ 1 = X µ s. Therefore X µ s t 1. Compare it with X µ σ N(0, 1). Example: Let X ad SX deote the sample mea ad sample variace of a idepedet radom sample of size 10 from a ormal distributio with mea µ = 0 ad variace σ. Fid c so that P X < c = SX 10

11 Defiitio: Let U χ 1 The F distributio ad V χ. If U ad V are idepedet the ratio U 1 V follows the F distributio with umerator d.f. 1 ad deomiator d.f.. We write X F 1,. The probability desity fuctio of X F 1, f(x) = Γ( 1+ ) Γ( 1 )Γ( Mea ad variace: Let X F 1,. The, ) ( 1 is: ) 1 ( x ) 1 1 x ( 1+ ), 0 < x <. E(X) =, ad var(x) = ( 1 + ) 1 ( ) ( 4). Shape: I geeral the F distributio is skewed to the right. The distributio of F 10,3 is show below: f(x) x 11

12 Applicatio: Let X 1, X,..., X i.i.d. radom variables from N(µ X, σ X ). Let Y 1, Y,..., Y m i.i.d. radom variables from N(µ Y, σ Y ). If X ad Y are idepedet the ratio S X σ X S Y σ Y F 1,m 1. Why? Example: Two idepedet samples of size 1 = 6, = 10 are take from two ormal populatios with equal variaces. Fid b such that P ( S 1 < b) = S 1

13 Distributio related to the ormal distributio χ, t, F - summary 1. The χ distributio: Let Z N(0, 1) the Z χ 1. Let Z 1, Z,, Z i.i.d. radom variables from N(0, 1). The Zi χ. Let X 1, X,, X i.i.d. radom variables from N(µ, σ). The ( X i µ σ ) χ. The distributio of the sample variace: ( 1)S σ χ 1, where S = 1 1 (x i x), x = 1 x i Let X χ, Y χ m. If X, Y are idepedet the X + Y χ +m.. The t distributio: Let Z N(0, 1) ad U χ. Z U t. Let X 1, X,, X i.i.d. radom variables from N(µ, σ). The x µ s t 1, where S = 1 (x i x) 1, x = 1 3. The F distributio: Let U χ ad V χ m. The U V m F,m with umerator d.f., m deomiator d.f. Let X 1, X,, X i.i.d. radom variables from N(µ X, σ X ) ad Let Y 1, Y,, Y m i.i.d. radom variables from N(µ Y, σ Y ) the: x i Useful: S X σ X S Y σ Y F 1,m 1 S X = 1 1 S Y = 1 m 1 t = F 1, where (x i x), x = 1 x i m (y i ȳ), ȳ = 1 m y i ad F α;,m = 1 F 1 α;m, 13

14 Practice questios Let Z 1, Z,, Z 16 be a radom sample of size 16 from the stadard ormal distributio N(0, 1). Let X 1, X,, X 64 be a radom sample of size 64 from the ormal distributio N(µ, 1). The two samples are idepedet. a. Fid P (Z 1 > ). b. Fid P ( 16 Z i > ). c. Fid P ( 16 Z i > 6.91). d. Let S be the sample variace of the first sample. Fid c such that P (S > c) = e. What is the distributio of Y, where Y = 16 Z i + 64 (X i µ)? f. Fid EY. g. Fid V ar(y ). h. Approximate P (Y > 105). i. Fid c such that 16 c Z i F 16,80. Y j. Let Q χ 60. Fid c such that ( ) Z1 P < c = Q k. Use the t table to fid the 80 th percetile of the F 1,30 distributio. l. Fid c such that P (F 60,0 > c) =

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