Continuous Density Functions

Question

First of all I know this question has been solved, but none of the online forums really explain it well, I am really confused with this question.

1. Suppose you choose at random a real number X from the interval $[2; 10]$.

   (a) Find the density function $f(x)$ and the probability of an event $E$ for this experiment, where $E$ is a subinterval $[a; b]$ of $[2; 10]$.

   (b) From (a), find the probability that $X > 5$, that $5 < X < 7$, and that $X^2 -12X + 35 > 0$.

For instance for part a, $f(x)= 1/8$. Why? If $X$ is chosen from the interval $[2;10]$, which means $2\le x\le 10$, from here there are nine numbers, so why is it $1/8$?

Answer 1

You are dealing with continuous random variable, not a discrete one.

The interval is a real number line segment of length $8$.   The uniform selection of REAL numbers from this interval will thus require a probability density of $\tfrac 1 8$, in order that this integral is unity.

$$\int_{[2;10]} f_X(x)\operatorname d x = \int_2^{10} \frac 1 8 \operatorname d x = 1$$

This leads into (b):

$$\mathsf P(a\leq X\leq b) = \int_a^b \tfrac 1 8\operatorname d x = \frac {b-a}{8} \qquad \mbox{iff }\Big[ 2\leq a\leq b\leq 10\Big]$$

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PS:

If $X$ is chosen from the interval $[2;10]$, which means $2≤x≤10$, from here there are nine numbers, so why is it $1/8$ ?

As mentioned above, you are dealing with a real number interval, so there are many, many, many more real numbers in the interval than those integers.   As to why the interval is of length $8$ when there are nine integers in the interval, refer to what is known as the "Fence Post Error".

Consider this: If I have nine fence posts placed in a line one metre apart, what length of fencing do I need?

Answer 2

From the first sentence we obtain that there is given a random variable $X:\Omega\to [2;10]$ such that for any Lebesgue measurable $A\subset[2;10]$ one has $$\mathbb{P}(X\in A)=\frac{\int_A1dx}{\int_{[2;10]}1dx}=\frac{|A|}{8},$$ where by $|A|$ I denote the one dimensional Lebesgue measure.

a) By definition a density function of a random variable $X$ is a Lebesgue measurable function defined on $X(\Omega)$ such that for $$\mathbb{P}(X\in A)=\int_Af(x)dx.$$

Getting back to our problem we look for $f$ such that $$\frac{|A|}{8}=\int_Af(x)dx.$$ Clearly $f(x)\equiv\frac{1}{8}$, satisfies the above condition. To answer the second part of a) we compute $$\mathbb{P}(X\in[a;b])=\int_{[a;b]}f(x)dx=\frac{b-a}{8}.$$

b) We have $$\mathbb{P}(X > 5)=\int_{(5;10]}f(x)dx=\frac{10-5}{8}=\frac{5}{8},\\ \mathbb{P}(5 < X < 7)=\int_{(5;7)}f(x)dx=\frac{7-5}{8}=\frac{1}{4}.\\ $$ To compute the last probability we first solve the inequality $x^2-12x+25>0$ for $x\in[2;10]$. The roots of the quadratic function are $x_{1,2}=6\pm\sqrt{6}$, so the solution is $x\in[2;6-\sqrt{6})\cup(6+\sqrt{6};10]$. Finally $$\mathbb{P}(X^2-12X+25>0)=\mathbb{P}(X\in[2;6-\sqrt{6})\cup(6+\sqrt{6};10])=\int_{[2;6-\sqrt{6})\cup(6+\sqrt{6};10]}f(x)dx=\frac{1}{8}(6-\sqrt{6}-2+10-(6+\sqrt{6}))=\frac{1}{8}(8-2\sqrt{6})=1-\frac{\sqrt{6}}{4}.$$