Let X1, X2, X10 follows standard normal distribution with meano and standard
deviation 1, then find the distribution of X, + X2 + ... + X10
Answers
Answer:
Let Z ∼ N(0,1). Then, if X = Z2, we say that X follows the chi-square distribution with 1 ... σ2. ∼ χ2 n−1. Proof: Example: Let X1,X2,...,X16 i.i.d. random variables from N(50,10). Find a.
Step-by-step explanation:
Distributions related to the normal distribution
Three important distributions:
• Chi-square (χ2
) distribution.
• t distribution.
• F distribution.
Before we discuss the χ2
, t, and F distributions here are few important things about the
gamma (Γ) distribution. The gamma distribution is useful in modeling skewed distributions
for variables that are not negative.
A random variable X is said to have a gamma distribution with parameters α, β if its
probability density function is given by
f(x) =
xα1ee
x
β
βαΓ(α)
, α, β > 0, x ≥ 0.
E(X) = αβ and σ2 = αβ2.
A brief note on the gamma function:
The quantity Γ(α) is known as the gamma function and it is equal to:
Γ(α) = Z ∞0
xα1e−x
dx.
Useful result:
Γ(
1
2
) = √
π.
If we set α = 1 and β =
1
λ
we get f(x) = λe−λx. We see that the exponential distribution is
a special case of the gamma distribution.
Theorem:
Let Z ∼ N(0, 1). Then, if X = Z
2
, we say that
X follows the chi-square distribution with 1
degree of freedom. We write,
X ∼
χ2
1.
Proof:
Find the distribution of X = Z2
, where f(z) = √12π ee 1
2 z2
. Begin with the cdf of X:
FX(x) = P(X ≤ x) = P(Z2 ≤ x) = P((√x ≤ Z ≤ √x) ⇒
FX(x) = FZ(
(
√x)
⇒
FZ(√x). Therefore:
fX(x) =
1
2
xx
1
2
1
√
2π
ee 1
2 x + 1
2
xx 1
2
1
√
2π
ee 1
2 x =
1
2 1
2
√πxx 1
2 ee x
2 , or
fX(x) =
x
x
1
2 e
e
x
2
2 1
2 Γ( 1
2 )
.
This is the pdf of Γ( 1
2 , 2), and it is called the chi-square distribution with 1 degree of freedom.
We write, X ∼ χ2
1.
The moment generating function of X ∼ χ2
1
is MX(t) = (1 ∞ 2t)− 1
2 .
Theorem:
Let Z1, Z2, . . . , Zn be independent random variables with Zi ∼ N(0, 1). If Y = P
n
i
=1
z2i
then
Y follows the chi-square distribution with n degrees of freedom. We write
Y ∼
χ
2
n
.
Proof:
Find the moment generating function of Y . Since Z1, Z2, . . . , Zn are independent,
MY (t) = MZ21 (t) × MZ22 (t) × . . . MZ
2
n
(t)
Each Z2i
follows χ2
1
and therefore it has mgf equal to (1 ∞ 2t)− 1
2 . Conclusion:
MY (t) = (1
∞
2t)
−
n
2 .
This is the mgf of Γ(
n
2 , 2), and it is called the chi-square distribution with n degrees of freedom.
Theorem:
Let X1, X2, . . . , Xn independent random variables with Xi ∼ N(µ, σ). It follows directly
form the previous theorem that if
Y = X
n
i=1
xi
i µ
σ
2
then Y ∼ χ2n.
