CS6015: Linear Algebra and Random Processes
Lecture 36: Uniform distribution, Normal distribution
Learning Objectives
What is the uniform distribution?
What is the normal distribution?
What are some properties of the normal distribution?
Uniform distribution (continuous)
Discrete Uniform Distribution
b−a1
\frac{1}{b - a}
b
b
pX(x)
p_X(x)
fX(x)
f_X(x)
Continuous Uniform Distribution
a1
a_1
b1
b_1
[
[
]
]
P(a1≤X≤b1)=b−a1(b1−a1)
P(a_1 \leq X \leq b_1) = \frac{1}{b-a} (b_1 - a_1)
a
a
b
b
…
\dots
…
\dots
b−a+11
\frac{1}{b - a + 1}
a
a
b
b
Expectation
fX(x)
f_X(x)
pX(x)
p_X(x)
a
a
b
b
…
\dots
…
\dots
b−a+11
\frac{1}{b - a + 1}
E[X]=∑xxpX(x)
E[X] = \sum_x xp_X(x)
E[g(X)]=∑xg(x)pX(x)
E[g(X)] = \sum_x g(x) p_X(x)
Var(X)=E[X2]−(E[X])2
Var(X) = E[X^2] - (E[X])^2
E[X]=∫−∞∞xfX(x)dx
E[X] = \int_{-\infty}^{\infty} xf_X(x) dx
E[g(X)]=∫−∞∞g(x)pX(x)dx
E[g(X)] = \int_{-\infty}^{\infty}g(x) p_X(x)dx
Var(X)=E[X2]−(E[X])2
Var(X) = E[X^2] - (E[X])^2
Mean and variance (Uniform Distribution)
E[X]=∫abxb−a1dx
E[X] = \int_{a}^{b} x\frac{1}{b-a} dx
E[g(X)]=∫−∞∞g(x)pX(x)dx
E[g(X)] = \int_{-\infty}^{\infty}g(x) p_X(x) dx
Var(X)=E[X2]−(E[X])2
Var(X) = E[X^2] - (E[X])^2
b−a1
\frac{1}{b - a}
b
b
fX(x)
f_X(x)
a
a
b
b
=2a+b
=\frac{a+b}{2}
E[X]=∫−∞∞xfX(x)dx
E[X] = \int_{-\infty}^{\infty} xf_X(x) dx
E[X2]=∫abx2b−a1dx
E[X^2] = \int_{a}^{b} x^2\frac{1}{b-a} dx
=b−a1(3b3−3a3)
=\frac{1}{b-a}(\frac{b^3}{3} - \frac{a^3}{3})
Var(X)=E[X2]−(E[X])2=12(b−a)2
Var(X) = E[X^2] - (E[X])^2 = \frac{(b-a)^2}{12}
E[X]=∫abxb−a1dx
E[X] = \int_{a}^{b} x\frac{1}{b-a} dx
E[X2]=∫abx2b−a1dx
E[X^2] = \int_{a}^{b} x^2\frac{1}{b-a} dx
E[X]=b−a1∗2x2∣ab
E[X] =\frac{1}{b-a} * \frac{x^2}{2} |_{a}^b
=b−a1∗(2b2−2a2)=b−a12b2−a2
= \frac{1}{b-a} * (\frac{b^2}{2} - \frac{a^2}{2}) = \frac{1}{b-a}\frac{b^2 - a^2}{2}
=b−a12(b−a)(b+a)=2a+b
= \frac{1}{b-a}\frac{(b - a)(b+a)}{2} = \frac{a+b}{2}
=b−a1∗3x3∣ab
=\frac{1}{b-a} * \frac{x^3}{3} |_{a}^b
=b−a1∗(3b3−3a3)=b−a13b3−a3
= \frac{1}{b-a} * (\frac{b^3}{3} - \frac{a^3}{3}) = \frac{1}{b-a}\frac{b^3 - a^3}{3}
=b−a13(b−a)(a2+ab+b2)=3a2+ab+b2
= \frac{1}{b-a}\frac{(b - a)(a^2 + ab + b^2)}{3} = \frac{a^2 + ab + b^2}{3}
Var(X)=E[X2]−E[X]2
Var(X) = E[X^2] - E[X]^2
=3a2+ab+b2−(2a+b)2
= \frac{a^2 + ab + b^2}{3} - (\frac{a+b}{2})^2
=3a2+ab+b2−4a2+2ab+b2
= \frac{a^2 + ab + b^2}{3} - \frac{a^2 + 2ab + b^2}{4}
=12(b−a)2
= \frac{(b-a)^2}{12}
Normal Distribution
Some fun with functions
x2
x^2
ex
e^x
x
x
e−x
e^{-x}
ex2
e^{x^2}
e−x2
e^{-x^2}
20∗e−x2
20 * e^{-x^2}
Some fun with functions
e−x2
e^{-x^2}
Normal distribution
N(0,1)=2π1e−x2
\mathcal{N}(0, 1) = \frac{1}{\sqrt{2 \pi}}e^{-x^2}
∫−∞∞2π1e−x2dx=1
\int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}}e^{-x^2} dx = 1
Normal distribution
N(0,1)=2π1e−x2
\mathcal{N}(0, 1) = \frac{1}{\sqrt{2 \pi}}e^{-x^2}
E[X]=∫−∞∞x⋅2π1e−x2dx
E[X] = \int_{-\infty}^{\infty}x\cdot\frac{1}{\sqrt{2 \pi}}e^{-x^2} dx
=0
= 0
Var[X]=E[X2]−(E[X])2
Var[X] = E[X^2] - (E[X])^2
=1
= 1
zero mean, unit variance
μ=0,σ=1
\mu = 0, \sigma = 1
Normal distribution (changing
N(μ,1)=2π1e−2(x−μ)2
\mathcal{N}(\mu, 1) = \frac{1}{\sqrt{2 \pi}}e^{-\frac{(x-\mu)^2}{2}}
E[X]=∫−∞∞x⋅2π1e−2(x−μ)2dx
E[X] = \int_{-\infty}^{\infty}x\cdot\frac{1}{\sqrt{2 \pi}}e^{-\frac{(x-\mu)^2}{2}} dx
=μ
= \mu
Var[X]=E[X2]−(E[X])2
Var[X] = E[X^2] - (E[X])^2
=1
= 1
μ=2
\mu = 2
μ=0
\mu = 0
μ=−2
\mu = -2
μ)
\mu)
Normal distribution (changing
N(0,σ2)=2π1e2σ2−x2
\mathcal{N}(0, \sigma^2) = \frac{1}{\sqrt{2 \pi}}e^{\frac{-x^2}{2\sigma^2}}
E[X]=∫−∞∞x⋅2π1e2σ2−x2dx
E[X] = \int_{-\infty}^{\infty}x\cdot\frac{1}{\sqrt{2 \pi}}e^{\frac{-x^2}{2\sigma^2}} dx
=0
= 0
Var[X]=E[X2]−(E[X])2
Var[X] = E[X^2] - (E[X])^2
=σ2
= \sigma^2
σ=0.5
\sigma = 0.5
σ=1
\sigma = 1
σ=2
\sigma = 2
σ)
\sigma)
Normal distribution (changing
N(μ,σ2)=2π1e2σ2−(x−μ)2
\mathcal{N}(\mu, \sigma^2) = \frac{1}{\sqrt{2 \pi}}e^{\frac{-(x-\mu)^2}{2\sigma^2}}
E[X]=∫−∞∞x⋅2π1e2σ2−(x−μ)2dx
E[X] = \int_{-\infty}^{\infty}x\cdot\frac{1}{\sqrt{2 \pi}}e^{\frac{-(x-\mu)^2}{2\sigma^2}} dx
=μ
= \mu
Var[X]=E[X2]−(E[X])2
Var[X] = E[X^2] - (E[X])^2
μ=2,σ=0.5
\mu = 2, \sigma = 0.5
μ,σ)
\mu, \sigma)
=σ2
= \sigma^2
The distribution is fully specified by the parameters:
μ,σ2
\mu, \sigma^2
Normal distribution
σ
\sigma
σ
\sigma
σ
\sigma
σ
\sigma
σ
\sigma
σ
\sigma
σ
\sigma
σ
\sigma
68%
68\%
95%
95\%
99%
99\%
P(μ−σ≤X≤μ+σ)
P(\mu - \sigma \leq X \leq \mu + \sigma)
=∫μ−σμ+σ2π1e2σ2−(x−μ)dx
= \int_{\mu - \sigma}^{\mu + \sigma} \frac{1}{\sqrt{2\pi}} e^{ \frac{-(x - \mu)}{2 \sigma^2} } dx
≈0.68
\approx 0.68
P(μ−2σ≤X≤μ+2σ)
P(\mu - 2\sigma \leq X \leq \mu + 2\sigma)
=∫μ−2σμ+2σ2π1e2σ2−(x−μ)dx
= \int_{\mu - 2\sigma}^{\mu + 2\sigma} \frac{1}{\sqrt{2\pi}} e^{ \frac{-(x - \mu)}{2 \sigma^2} } dx
≈0.95
\approx 0.95
Learning Objectives
What is the uniform distribution?
What is the normal distribution?
What are some properties of the normal distribution?
CS6015: Linear Algebra and Random Processes Lecture 36: Uniform distribution, Normal distribution
CS6015: Lecture 36
By Mitesh Khapra
CS6015: Lecture 36
Lecture 36: Uniform distribution, normal distribution
- 1,893
