Consider $\Omega=[0,1]$ with probability defined on intervals by $P([a,b])=b-a, 0\le a\le b\le1$

Question

Consider $\Omega=[0,1]$ with probability defined on intervals by $P([a,b])=b-a, 0\le a\le b\le1$

I have tried to explain this sum to the best of my ability. To summarize, this could have been a simpler sum, had the random variable not been a minimum function.

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Defined as a random variable by: $X(\omega)=\min\{\omega,1-\omega\}.$

Please find attached an image/link to the figure I have drawn for the described random variable. enter image description here

a) Find

$P(\{\omega:X(\omega)\le\frac{4}{5}\})$,

$P(\{\omega:X(\omega)\le\frac{1}{4}\})$,

$P(\{\omega:X(\omega)\le0\})$,

$P(\{\omega:X(\omega)\le-1\})$.

b) Find $F_{X}(a)$ for all $a\in\mathbb{R}$. That is, find the probability that X is less or equal to a

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I started the sum with;
$X(\omega)=\min\{\omega,1-\omega\}={\omega,.if. \omega\leqq\frac12}$
$;.{1-\omega,.if.\omega>\frac12}$

However, I am unable to calculate the probabilities asked for the given values of the random variable.
Where can I be missing to check? How can we find the probability of a single number (as asked), when we have been given a minimum function.

Answer 1

I think it is easier to work visually on this. Draw horizontal lines on the image corresponding to the $\alpha$ in $P[X \le \alpha]$.

There are three cases to consider:

(i) $\alpha \ge {1 \over 2}$, in which case $P[X \le \alpha] = 1$.

(i) $\alpha < 0$, in which case $P[X \le \alpha] = 0$.

(iii) $\alpha \in [0,{1 \over 2})$, in which case $P[X \le \alpha] = 2 \alpha$.

To see that latter, note that $\{ \omega \in [0,1] | X(\omega) \le \alpha \} = [0, \alpha] \cup [1-\alpha,1]$.

Answer 2

Use your graph!

If you want to know, for example, when is $X(\omega)\le 0.2$, draw a horizontal line at height $0.2$ (think that the horizontal axis stands for $\omega$ and the vertical for $X$). You will see that $X(\omega)$ is $0.2$ or less (the graph is below the horizontal line) when $\omega\in [0,0.2]\cup [0.8,1]$.

So $P\big(X(\omega)\le 0.2\big)=P\big([0,0.2]\cup[0.8,1]\big)$ and then use the definition of $P$.

The same graph will make evident that $X(\omega)\le a$ is equivalent to $\omega \in [0,1]$ for $a\ge\tfrac12$, and is equivalent to $\emptyset$ for $a<0$.