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)
F(x) = \begin{cases}<br /> 0 &amp;\text{if} \;x \leq 1 \\<br /> \displaystyle 1 - \frac{1}{x^2}&amp; \text{if} \;x &gt; 1<br /> \end{cases}<br />
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
P(X = x) = P(X \leq x) - P(X &lt; x) = F(x) - F(x-0).
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:
F_X(x) = \begin{cases} 1 - e^{-x/3}, &amp; x &lt; 5 \\<br /> 1, &amp; x \geq 5<br /> \end{cases}
Here, $P(X = 5) = e^{-5/3} = F_X(5) - F_X(5-0)$. Other examples abound.

 

[toothpaste666]

Thanks for clearing that up :)