CDF 概率计算

Criado em: 24 de novembro de 2024

Respondido usando O1 Preview por Chat01

Pergunta

Suppose that the c.d.f. FF of a random variable XX is as sketched in Fig. 3.9. Find each of the following probabilities:

a. Pr(X=1)\Pr(X = -1)
b. Pr(X<0)\Pr(X < 0)
c. Pr(X0)\Pr(X \leq 0)
d. Pr(X=1)\Pr(X = 1)
e. Pr(0<X3)\Pr(0 < X \leq 3)
f. Pr(0<X<3)\Pr(0 < X < 3)
g. Pr(0X3)\Pr(0 \leq X \leq 3)
h. Pr(1<X2)\Pr(1 < X \leq 2)
i. Pr(1X2)\Pr(1 \leq X \leq 2)
j. Pr(X>5)\Pr(X > 5)
k. Pr(X5)\Pr(X \geq 5)
l. Pr(3X4)\Pr(3 \leq X \leq 4)
这个图表展示了累积分布函数(c.d.f.)F(x)F(x)的一个示例,具有以下特点:

  • 对于 x<1x < -1F(x)=0F(x) = 0
  • x=1x = -1 处有一个跳跃,F(x)F(x) 突然跃升至 0.2。
  • 0x10 \leq x \leq 1 范围内,F(x)F(x) 保持恒定为 0.2。
  • x=1x = 1 开始,F(x)F(x) 线性增加,到 x=3x = 3 达到 0.6。
  • x=3x = 3 处函数保持恒定,随后在 x=4x = 4 处再次缓慢增加。
  • x=5x = 5 时,函数达到最大值 1,并保持不变。

这是一个典型的分段连续分布示意图。

Pensamento

Reconstructing the CDF

I'm carefully plotting the cumulative distribution function (CDF) F(x) based on the user’s description, identifying jumps and constant segments, and ensuring continuity and piecewise definition.

Mapping out behavior

OK, let me see: For x<1x < -1, F(x)=0F(x) = 0. At x=1x = -1, F(x)F(x) jumps to 0.2, and it stays 0.2 for 1<x<0-1 < x < 0.

Analyzing function dynamics

It's interesting to see that from x=1x = -1 to x=0.2x = 0.2, F(x)F(x) stays steady at 0.2. Then from x=1x = 1 to x=3x = 3, it increases linearly to 0.6 and stays constant afterward until x=5x = 5, where it jumps to 1.

Mapping out the CDF

The CDF F(x)F(x) is defined piecewise:

  • F(x)=0F(x) = 0 for x<1x < -1,
  • 0.20.2 for 1x1-1 \leq x \leq 1,
  • 0.2+0.2(x1)0.2 + 0.2(x - 1) for 1<x<31 < x < 3,
  • 0.60.6 for x=3x = 3,
  • 0.60.6 for 3<x43 < x \leq 4 (constant),
  • 0.6+0.4(x4)0.6 + 0.4(x - 4) for 4<x54 < x \leq 5,
  • 11 for x5x \geq 5.

Mapping the increase

I’m thinking of F(x) as a piecewise function, increasing linearly from 0.6 to 1 between x=4x=4 and x=5x=5, with different expressions for other ranges.

Determining discrete probabilities

I'm exploring discrete probabilities using P(X=x)=F(x)limtxF(t)P(X = x) = F(x) - \lim_{t \to x^-} F(t). Notably, at x=1x = -1, there's a jump from F(1)=0F(-1^-) = 0 to F(1)=0.2F(-1) = 0.2, indicating a discrete probability scenario.

Assessing probabilities
I’m determining Pr(X=1)=0.2\Pr(X = -1) = 0.2 by examining the jump size at x=1x = -1. For Pr(X<0)\Pr(X < 0) and Pr(X0)\Pr(X \leq 0), I evaluate the cumulative distribution function. Finally, I’m checking for the jump at x=1x = 1.

Breaking down probabilities

I’m looking at the probability calculations for different ranges. For 0<X30 < X \leq 3, I subtract F(0)F(0) from F(3)F(3) to get 0.4. For 0<X<30 < X < 3, I just use F(3)F(0)F(3^-) - F(0) since there’s no jump at x=3x = 3.

Confirming the probability

Next, I observed that since F(0)F(0^-) equals F(0)F(0) at 0, and there is no jump at x=0x=0, Pr(0X3)\Pr(0 \leq X \leq 3) consistently results in 0.4, which aligns with previous calculations.

Evaluating probabilities

I’m calculating specific probabilities given the cumulative distribution function F(x)F(x). For Pr(0X3)\Pr(0 \leq X \leq 3), it’s F(3)F(0)=0.4F(3) - F(0^-) = 0.4. For Pr(1<X2)\Pr(1 < X \leq 2), I’m determining F(2)F(1)F(2) - F(1) using the linear trend in the distribution.

Confirming the calculations

I calculated F(2)F(2) and F(1)F(1) to determine Pr(1<X2)\Pr(1 < X \leq 2), confirming no jump at x=1x=1.

Assessing probabilities

I’m calculating Pr(1X2)\Pr(1 \leq X \leq 2) and Pr(X>5)\Pr(X > 5) using cumulative distribution function (CDF) values. Interestingly enough, there’s no jump at x=1x=1 and x=5x=5.

Calculating probabilities

I’m curious about the probability for different ranges of XX. For Pr(X5)\Pr(X \geq 5), since F(5)=1F(5^-) = 1 and there's no jump at x=5x=5, the probability is 00. For Pr(3X4)\Pr(3 \leq X \leq 4), F(x)F(x) is constant at 0.60.6 from x=3x=3 to x=4x=4, so the probability remains 0.60.6.

Resposta

To solve these probability problems, we'll interpret the given cumulative distribution function (CDF) F(x)F(x) based on the description provided. We'll use the properties of CDFs and the fact that probabilities can be found using the differences in the CDF values.

First, let's reconstruct F(x)F(x) based on the description:

  1. For x<1x < -1: F(x)=0F(x) = 0.
  2. At x=1x = -1: There's a jump to F(1)=0.2F(-1) = 0.2.
  3. From 1<x1-1 < x \leq 1: F(x)=0.2F(x) = 0.2 (constant).
  4. From 1<x<31 < x < 3: F(x)F(x) increases linearly from 0.2 to 0.6.
  5. From 3x43 \leq x \leq 4: F(x)=0.6F(x) = 0.6 (constant).
  6. From 4<x<54 < x < 5: F(x)F(x) increases linearly from 0.6 to 1.
  7. For x5x \geq 5: F(x)=1F(x) = 1.

Now, let's compute each probability:

a. Pr(X=1)\Pr(X = -1)

  • There's a jump at x=1x = -1 from 0 to 0.2.
  • Answer: Pr(X=1)=F(1)F(1)=0.20=0.2\Pr(X = -1) = F(-1) - F(-1^-) = 0.2 - 0 = \boxed{0.2}.

b. Pr(X<0)\Pr(X < 0)

  • Pr(X<0)=F(0)=0.2\Pr(X < 0) = F(0^-) = 0.2.
  • Answer: 0.2\boxed{0.2}.

c. Pr(X0)\Pr(X \leq 0)

  • Pr(X0)=F(0)=0.2\Pr(X \leq 0) = F(0) = 0.2.
  • Answer: 0.2\boxed{0.2}.

d. Pr(X=1)\Pr(X = 1)

  • No jump at x=1x = 1.
  • Answer: Pr(X=1)=F(1)F(1)=0.20.2=0\Pr(X = 1) = F(1) - F(1^-) = 0.2 - 0.2 = \boxed{0}.

e. Pr(0<X3)\Pr(0 < X \leq 3)

  • Pr(0<X3)=F(3)F(0)=0.60.2=0.4\Pr(0 < X \leq 3) = F(3) - F(0) = 0.6 - 0.2 = \boxed{0.4}.

f. Pr(0<X<3)\Pr(0 < X < 3)

  • No jump at x=3x = 3, so same as (e).
  • Answer: 0.4\boxed{0.4}.

g. Pr(0X3)\Pr(0 \leq X \leq 3)

  • No jump at x=0x = 0, so same as (e) and (f).
  • Answer: 0.4\boxed{0.4}.

h. Pr(1<X2)\Pr(1 < X \leq 2)

  • Compute F(2)=0.2+0.2(21)=0.4F(2) = 0.2 + 0.2(2 - 1) = 0.4.
  • Pr(1<X2)=F(2)F(1)=0.40.2=0.2\Pr(1 < X \leq 2) = F(2) - F(1) = 0.4 - 0.2 = \boxed{0.2}.

i. Pr(1X2)\Pr(1 \leq X \leq 2)

  • No jump at x=1x = 1, so same as (h).
  • Answer: 0.2\boxed{0.2}.

j. Pr(X>5)\Pr(X > 5)

  • F(5)=1F(5) = 1, so Pr(X>5)=1F(5)=0\Pr(X > 5) = 1 - F(5) = 0.
  • Answer: 0\boxed{0}.

k. Pr(X5)\Pr(X \geq 5)

  • No jump at x=5x = 5, so Pr(X5)=1F(5)=0\Pr(X \geq 5) = 1 - F(5^-) = 0.
  • Answer: 0\boxed{0}.

l. Pr(3X4)\Pr(3 \leq X \leq 4)

  • F(x)F(x) is constant between 3 and 4.
  • Answer: Pr(3X4)=F(4)F(3)=0.60.6=0\Pr(3 \leq X \leq 4) = F(4) - F(3^-) = 0.6 - 0.6 = \boxed{0}.

Summary of answers:

a) 0.2

b) 0.2

c) 0.2

d) 0

e) 0.4

f) 0.4

g) 0.4

h) 0.2

i) 0.2

j) 0

k) 0

l) 0

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