Let X denote the time to failure (in years) of a hydraulic component. Suppose the pdf of X is f(

Let X denote the time to failure (in years) of a hydraulic...

Key Concepts

Interval Probability Calculation

For continuous random variables, the probability that the variable falls within a certain interval can be computed by evaluating the difference between the cdf at the upper bound and the cdf at the lower bound of the interval. This method efficiently simplifies the process of finding probabilities over intervals.

Legitimacy of a Probability Density Function

A probability density function (pdf) must be nonnegative over its entire domain and the integral of the pdf over that domain must equal 1. Verifying that a function qualifies as a pdf often requires computing an integral, which, in the case of unbounded or complicated domains, may be an improper integral.

Cumulative Distribution Function (cdf)

The cumulative distribution function (cdf) of a continuous random variable is defined as the integral of the pdf from the lower bound of the variable’s domain up to a specified value. The cdf represents the probability that the random variable takes on a value less than or equal to that specified value.

Expected Value of a Random Variable

The expected value (or mean) of a continuous random variable is computed as the integral of the product of the variable and its pdf over the entire range of the variable. This measure provides an estimate of the long-run average outcome for the random process.

Expected Value of a Function of a Random Variable

When interested in the average outcome of a function applied to a random variable, the expected value is computed by integrating the product of that function and the pdf over the variable’s domain. This approach is essential for evaluating transformed random variables, such as calculating the expected salvage value in reliability or financial contexts.

Transcript

00:02 Okay, so we've been given a pdf for a function that is equal to 32, so we're going to say the f of x, which is the pdf, is equal to 32 divided by x plus 4, it's cute, for all x greater than zero.

00:21 So this is our function right here, our problem statement right here, and the first thing we're going to do is we're going to verify that this pdf, this, this f of x here, is legitimate.

00:30 And what that means is that the integral from negative infinity to infinity of this pdf is equal to 1.

00:35 However, given this boundary condition here, we are going to instead swap the lower bound for 0 instead of negative infinity.

00:42 So we're going to take this integral, integral right there, it's lagging out a little bit, there we go, of f of x from zero to infinity.

00:51 This is dx.

00:53 Which when we evaluate with this expression here, we'll give us an integral, which i will save some time here by not performing, you know, all the little.

01:02 Steps, but we're going to get negative 16 divided by x plus 4.

01:11 And this is our integral right there.

01:14 And now we're going to evaluate it at our bounds, which are zero and infinity.

01:18 And we plug this in, we can think about this a little bit.

01:21 So we plug in infinity here, and we take the limit, right, if you're being mathematically accurate.

01:25 So we're going to take the limit as x approaches infinity.

01:27 Well, this term is all going to go to zero.

01:29 So the first term of this substitution of our bounds is going to be zero.

01:35 And now for the second term, when we plug in this zero here, we're going to see that this x is zero, so just 16, or negative 16 over 16.

01:44 So that's going to be negative 16 divided by 16, which when you evaluate this whole expression here, this is going to be equal to 1, which is what we're looking for, right? so what this tells us is that this pdf is a legitimate pdf, but how we can just think about it a little bit is that when we added up every possible value of x that this pdf can take, they all added it.

02:05 To 1, or they all add it to 100%.

02:07 There's another way to look at it.

02:08 So we know that all the probabilities add some to 100%, which is good, which is what we're looking for.

02:15 From here, the second thing we need to do, because we're not quite done verifying that this is legitimate, is we need to verify that f of x is greater than 0 for all x, which in this case is for all x greater than 0.

02:26 So we're going to check this function right here.

02:28 We're just going to look at it for a second.

02:29 Sorry, i cut it off at the top.

02:30 We're going to look at it for a second.

02:31 We're going to say, ok.

02:32 So for any value of x, i pick, what's this going to look like.

02:37 And when we see is that, okay, so for all positive x, which is the range that we're, the domain that we're restricted to, for all positive x, this function is going to be positive.

02:46 So that satisfies our second check.

02:48 So we can say that this pdf is a legitimate pdf.

02:51 From there, we can move on to the rest of the problems.

02:55 So we're going to say that the cdf is the integral of the pdf.

02:59 And it's kind of handy that we already have that written right here, except we have some different bounds.

03:04 So for the cdf, we're going to integrate from zero but to x because we don't know the value that we want to take this accumulation up to.

03:12 It's because we're adding all the probabilities up to some value, right? so we're going to take it from zero to x of the pdf, which is f of x, that's a little f, dx, which is handy because we happen to already have this, but this exact calculation or the solution to this right here...

Please give Ace some feedback