Probability Density Function - Is my solution correct? Need help with a limit of an integral

Question

I have no answers to refer to, hence, would be great if someone could check up if my procedure to solve the following problem is correct. Also, I am struggling to solve for event B from part b) - any tips would be much much appreciated! I'm preparing for an exam, hence, it is of vital importance. Thank you!

Consider the following function $f$ on $I=[2,∞): f(x)=ax^{-3}$ where $a$ is a certain constant.

a) Find $a$ such that $f$ is a probability density function.

By rule it must hold that: $P(I)=\int_If(x)dx=1$

From here follows:

$P(I)=\int_If(x)dx=\int^∞_2ax^{-3}dx=[-\cfrac{a}{2}x^{-2}]^{x=∞}_{x=2}= > [-\cfrac{a}{2}x^{-2} - [-\cfrac{a2^{-2}} {2}]]$ The first part goes to 0 as x goes to ∞. So we are left with the second part, so, $\cfrac{a2^{-2}} {2}=1 => a=8$

b) Determine the probabilities of the events $A=(4,∞)$ and $B=${3}

$P[4,∞]=\int^∞_48x^{-3}=[-4x^{-2}]^{x=>∞}_{x=4}=[-4x^{-2} - > [-4*4^{-2}]$ Once again, as x goes to infinity, the first part goes to 0 so we are left with the second part only, so $4*4^{-2}=1/4$

For event B, though, I have absolutely no idea what to do. What limits do we put on the integral, both upper and lower limit 3? But that would lead us to an equation that equals 0 so I'm confused.

Answer 1

The work you have done is correct.

Regarding the event B from part b) what you did is also correct. When dealing with continuous probability every single $x$ has probability $P(X = x) = 0$. The reason is exactly as you said: lower and upper bounds of the integral are the same so the integral is equal to $0$.

If you are confused about probability of event being $0$ think about it this way. Imagine that you are slicing a stick at a random position. To do so you must pick a point at random from interval $(0, L)$ where $L$ is the length of a stick. Since there are infinitely many points on a stick, probability of picking specific one has to be $0$.

It only makes sense to have positive probability for a range (interval) of points. For example let $a, b$ ($b > a$) be points on a stick with length $L$. Now the probability that you pick a point between $a$ and $b$ is

$$P(X > a, X < b) = \frac{b-a}{L}$$

If each point had a positive probability and there are infinitely many points in any real interval your total probability would be $\infty$ which contradicts the rule that total probability must be equal to $1$.

Answer 2

Your procedures for finding $\ a\ $ and the probability of the event $\ A\ $ are correct, and I get the same answers that you did.

The answer to your question about finding the probability of $\ B\ $ is yes indeed, you do put 3 as both the upper and lower limit of the integral, and the probability of the event $\ B\ $. For continuous distributions (which include all those with density functions) the probability of any event containing at most a countable set of sample points is always zero.