# Trying to understand random variables

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1. Sep 18, 2016

1. The problem statement, all variables and given/known data

Let's say you have a number from [-2,4], with X(ζ) = -ζ + 4

Find (a) P([-2,4]) and (b) P({X≤2})

2. Relevant equations
{X = x} = {ζ ∈ S: X(ζ) =x }

3. The attempt at a solution

It looks like my sample space, S = [-2,4].

(a) For P([-2,4])

{-2 ≤ X ≤ 4} = {ζ ∈ S: -2 ≤ X(ζ) ≤ 4}

= {ζ ∈ S: -2 ≤ -ζ + 4 ≤ 4}

= {ζ ∈ S: 0 ≤ ζ ≤ 6}

= [0,6], but since ζ ∈ S => [0,4]

At this point I didn't know how I find my probability, do I take the quotient of the (length)/(length of my S)

I.e. P[(-2,4)] = 4/6?

(b) If this is true, does that mean:

P({X≤2}) = {X ≤ 2} = {ζ ∈ S: X(ζ) ≤ 2}

= {ζ ∈ S: -ζ + 4 ≤ 2}

= {ζ ∈ S: 2 ≤ ζ } = [2]

= 1/6

Any advice and tips would be appreciated

Last edited: Sep 18, 2016
2. Sep 18, 2016

### andrewkirk

To understand a real-valued random variable $X$ we need at least four things:

1. A sample space $\Omega$. You have said this is the interval [-2,4]
2. A sigma field $\Sigma$ that tells us which subsets of $\Omega$ are measurable. This is not specified in your problem but that's OK because it's usual to assume in such cases that it is the collection of Borel sets, which is basically any reasonably well-behaved set.
3. A probability measure $P$ that assigns probabilities to subsets of the sample space that are in $\Sigma$. THis is missing from the problem statement.
4. A function $X$ from $\Omega$ to $\mathbb R$. You have given this as $\zeta\mapsto 4-\zeta$.

The problem statement is incomplete because $P$ is not given.

Part (a) can be solved because it just asks the probability measure of the entire sample space, which by definition is always 1.

Part (b) asks for the measure of the following subset of $\Omega$:
$$\{\zeta\in\Omega\ :\ -\zeta+4\leq 2\}$$
Since this is a proper subset of $\Omega$ and we have not been told the probability measure $P$, the problem cannot be solved.

Are you sure you haven't left anything out of your recounting of the problem? Did they for instance say that $X$ is uniformly distributed?

Also, the calculations in (b) have an error. $X\leq 2$ is not the same as $-2\leq X\leq 2$. What is the range of $X$?

3. Sep 18, 2016

I just took a look at the problem statement, it does not mention anything about X being uniformly distributed. I fixed the range for part (b). Are real valued random variable problems usually assumed to be uniformly distributed?

4. Sep 18, 2016

### andrewkirk

Not necessarily. It would help if you posted an image - eg a clear photo - of the problem statement exactly as given to you.

5. Sep 18, 2016

Unfortunately everything I posted is already shown in the original post.

"Choose a number a random from [-2,4], and let ܺX(ζ) = -ζ + 4. Find the following:

(a)P([-2,4])
(b) P({X≤2})
"

6. Sep 18, 2016

### micromass

Staff Emeritus
That is insufficient information to solve the problem.

7. Sep 18, 2016

### LCKurtz

No, I'm sure it did not state that. Perhaps it stated "Choose a number at random from...", which might plausibly be interpreted as being from a uniform distribution.

8. Sep 19, 2016

### Ray Vickson

It is not clear whether (a) asks for $P\{ \zeta \in [-2,4] \}$ (which, of course, equals 1 by definition of the sample space for $\zeta$), or whether it asks for $P\{ X(\zeta) \in [-2,4]\}$ (which, of course, is the same as $P\{ X(\zeta) \in [0,4]\} = P\{ \zeta \in [0,6] \} = P\{ \zeta \in [0,4] \}$ as you said).
Perhaps part (b) just wants you to express $P \{ X \leq 2 \}$ in terms of the (cumulative) distribution function $F_{\zeta}(z)$ of the random variable $\zeta$.

9. Sep 20, 2016

### haruspex

Yes, I'd say it is making the common misuse of "at random" to imply a uniform distribution. Strictly speaking, it tells you nothing about the distribution, it could even be deterministic. One hopes for better from textbooks and teachers.