Finding Probabilities for a Probability Density Function | Homework Solution

[toothpaste666]

Homework Statement

If the distribution function of a random variable is given by
F(x) = 1- 1/x^2 for x>1
and
F(x) = 0 for x <= 1

find the probabilities that this random variable will take on a value
a) less than 3
b) between 4 and 5

The Attempt at a Solution

since they use the capital F i think that means it is a cumulative probability distribution. So in order to find part a) we need to find the cumulative probability that it takes on a value from 3 to infinity and subtract this from 1
the antiderivative of the function for x>1 is

x + 1/x

we need to evaluate this from x=3 to x = infinity
we have
∞ + 1/∞ - (3 + 1/3)
= ∞ + 0 -3 - 1/3
= ∞ - 3.333

I know this can't be right because of the ∞ but I am not sure what I am doing wrong =[ I would appreciate it if someone can't point me in the right direction

 

[toothpaste666]

wait a minute. I found in my book that dF(x)/dx = f(x)
so since F(x) = 1 - 1/x^2

f(x) = dF(x)/dx = d/dx(1-1/x^2) = 0 - d/dx(x^-2) = -(-2x^-3) = 2x^-3

so to find the cumulative probability it takes on a value from 3 to ∞ we just plug in
(1- 1/∞^2) - (1-1/3^2)
= (1-0) - (1- 1/9)
= 1 - .888888...
= .111111...
but since we want the probability of the value taking on a value less than 3, we subtract this from 1
so the answer to part a) is .888888
is this right?

 

[Ray Vickson]

wait a minute. I found in my book that dF(x)/dx = f(x)
so since F(x) = 1 - 1/x^2

f(x) = dF(x)/dx = d/dx(1-1/x^2) = 0 - d/dx(x^-2) = -(-2x^-3) = 2x^-3

so to find the cumulative probability it takes on a value from 3 to ∞ we just plug in
(1- 1/∞^2) - (1-1/3^2)
= (1-0) - (1- 1/9)
= 1 - .888888...
= .111111...
but since we want the probability of the value taking on a value less than 3, we subtract this from 1
so the answer to part a) is .888888
is this right?

Why are you doing it the hard way? You are given the distribution function (= cumulative distribution)
[tex] F(x) = \begin{cases}
0 &\text{if} \;x \leq 1 \\
\displaystyle 1 - \frac{1}{x^2}& \text{if} \;x > 1
\end{cases}
[/tex]
This is ##P(X \leq x)## already, so no more work is needed: ##P(X < 3) = P(X \leq 3) = 1 - 1/9 = 8/9##.

Incidentally, saying that the function must be a cumulative distribution just because it is denoted by "F" is about the worst justification you could possibly offer. It is a (cumulative) distribution function because it is monotone non-decreasing, and has limits of 0 at -∞ and 1 at +∞. Those are essentially the defining properties of a (cumulative) distribution. Besides that, the problem called it a distribution function, and modern usage leans towards omission of the adjective "cumulative" (so that distribution = cumulative distribution often, nowadays).

 

[toothpaste666]

For the continuous probability distributions, will there ever be a case where P(X<3) is not equivalent to P(X<=3)? or will I always be able to plug it right in like that?

 

[Ray Vickson]

For the continuous probability distributions, will there ever be a case where P(X<3) is not equivalent to P(X<=3)? or will I always be able to plug it right in like that?

If the (cumulative) distribution function ##F(x)## is continuous, then ##P(X \leq x = P(X < x)## for every ##x##.

Differences come in when the function ##F(x)## has jump discontinuities; this occurs in so-called "mixed" distributions, which describe random variables that are partially continuous and partly discrete. In such cases, the probability of a single point ##P(X = x)## need not be zero anymore, and we have
[tex] P(X = x) = P(X \leq x) - P(X < x) = F(x) - F(x-0). [/tex]
Here, we have adopted the customary convention that ##F(x) = P(X \leq x)## is a right-continuous function (that is, ##\lim_{ y \downarrow x} F(y) = F(x) ##), and ##F(x-0)## is the left-hand limit at ##x##; that is, ##F(x-0) = \lim_{y \uparrow x} F(y)##.

Such cases are NOT unusual or pathological. They occur, for example, when you truncate a random variable to obtain another one, or when the random variable describes, say, an equipment lifetime that may be ##X = 0## if you bought a "lemon", but is otherwise a continuous random variable in the region ##\{ x > 0 \}##. An example of a truncated random variable might be the lifetime of a piece of equipment that we will scrap for sure if it reaches age = 5 years; otherwise, the lifetime might, for example, be exponential with mean 3 years. In this case, the lifetime ##X## would have a cdf with a jump discontinuity at x = 5:
[tex] F_X(x) = \begin{cases} 1 - e^{-x/3}, & x < 5 \\
1, & x \geq 5
\end{cases} [/tex]
Here, ##P(X = 5) = e^{-5/3} = F_X(5) - F_X(5-0)##. Other examples abound.

 

[toothpaste666]

Thanks for clearing that up :)