Enginerring Stats question... Integration related

urahonky

So I'm taking my STT363 class and I've run into a slight bump. I forgot my calc notes as I've thrown them away in a fiery pit.

Anyway. The question (not the hw one, the one above it) says:

The pdf for Y is f(y) = k sqrt(y), for 0<y<1. Where K is a constant.

Okay, so basically f(y) is the kth root of y. The first question asks: "Find the value of k, so f(y) is a legitamite pdf". I assume it is asking find out what k's make f(y) valid? That would be k>0 right? I feel like this is wrong.

Also, i forget how to integrate sqrt(x), can anyone give me a refresher? I love you guys.

Terrendos

I'm not sure what a "pdf" is. Can you not use the acronym? It might ring a bell.

d/dx of root of x (remember root x is x to the one half power) is one half times x to the negative one half power, or 1/(2*X^1/2).

urahonky

I'm sorry Terrendos, it stands for "probability density function".

physi_marc

The function is kth root of y? Are you sure? Isn't K just a constant in front of the square root? In that case, they're asking you to normalize f(y), that is, if you integrate f(y)dy over 0 to 1, it must be equal to 1 since it's a PDF. Does that make sense to you?

AgentBryant

I'm making the assumption that the OP is just at a loss of how to type the kth root of y. In that case, f(y) = y^(1/k)

Integrating this from 0 to 1 must equal 1 in order for this to be a PDF.

urahonky

AgentBryant wrote: »

I'm making the assumption that the OP is just at a loss of how to type the kth root of y. In that case, f(y) = y^(1/k)

Integrating this from 0 to 1 must equal 1 in order for this to be a PDF.

Bryant has it correct. Sorry, typing kth root of y had me at a loss. :P

I see what you're saying, but how do you integrate y^(1/k)?

Integrating y^(1/2) is 3y^(3/2)/2 right? Terrendos I think you were going backwards when integrating... It's sad that I don't remember what the opposite of integrating is...

KafkaAU

Differentiation.

You said k was a constant, so you treat it as a constant.

i.e. the integral of y^1/k = ((1/k)+1)y^((1/k)+1)

Or maybe more easily (1+k)/k * y^((1+k)/k)

And to finish off:

(1+k)/k*(1)^((1+k)/k) - (1+k)/k*(0)^((1+k)/k)

(1+k)/k

So to be a legitamate PDF (1+k)/k = 1

Which is impossible, so no its not.

AgentBryant

A little imagery for clarity:



And of course the definite integral from 0 to 1 should equal 1 for f(y) to be a PDF, since the integral is really the summation of the probabilities, which sum to 1. The rest would be using algebra to find a k where f(y) is a PDF.

urahonky

Aha, I suppose that makes sense. Thanks KafkaAU I really do appreciate it.

e: And pictures help too, thanks Bryant. I appreciate the responses. Time to finish up this stupid Stats homework.

Folken Fanel

Hey guys, phd statistics student and once-teacher of engineering stats here. All pdfs integrate to 1. You could try integrating the function, set it equal to 1 and then solve for k. This won't help you though because you can't isolate k. Are you sure you are writing the question correctly?

EDIT: Gah, right plug in 0 and 1 and solve for k.

EDIT2: Be careful though... you'll get k = 0. Which actually makes sense since y^0 = 1 is a valid pdf on (0,1), i.e. the standard uniform distribution.