Probability Density (and computing constant K)

[Predz]

Homework Statement

The random variable X has the probability density f(x) defined by:

f(x) = ke^(-|x|)

Compute the constant k, then find the probability density of Y=X²

Homework Equations

The Attempt at a Solution

I'm completely stumped by this question. I know I need to integrate ke^(-|x|) to find k, but after I integrate, what values do I replace x by to find k? I also have no idea what to do for the probability density part...

Any tips are welcomed. Please help. :(

 

[Predz]

So I worked out the integration part by myself. When I integrated ke^(-|x|), I got -ke^(-|x|). To find k, I substituted x by 0 and -\infty. I worked it out and I found out that k=-1. I'm not sure I did this right though...

 

[LPerrott]

What conditions does a function need to satisfy in order to be a probability density function? Hint: There are two of them, one is that f(x)>=0, but it is the the second condition that will help you find your answer.

 

[Predz]

What conditions does a function need to satisfy in order to be a probability density function? Hint: There are two of them, one is that f(x)>=0, but it is the the second condition that will help you find your answer.

Is it negative infinity? See the post above yours.

 

[LPerrott]

No. You need to look up the properties of a probability density function. Simply integrating the prob dens function isn't enough to find the value of k. If I want k = __ , then I need to start off with an equation integral of (___) = ___

 

[Predz]

No. You need to look up the properties of a probability density function. Simply integrating the prob dens function isn't enough to find the value of k. If I want k = __ , then I need to start off with an equation integral of (___) = ___

Has to be equal to 1.

So I integrated ke^(-|x|) and got -ke^(-|x|). Then I did -ke^(-|x|) = 1 and replaced x by 0 and negative infinity. I solved and got k = -1. Is my integration correct? Was I right to replace x by 0 and negative infinity? Am I completely off?

 

[LPerrott]

I can tell you without even checking integration that is incorrect. k=-1 violates the first property of a prob density function ... mainly that f(x)>=0. Notice you have an absolute value of x in the function. This might be what's confusing you. Try solving the following equation for k

2k int(0 to infty)(e^-x dx) = 1

Sorry i don't know how to insert Latex into my reply. In words, you should use the equation 2 times k times the integral from 0 to infinity of e^-x equals 1

 

[Predz]

I re-did it again (this time using infinity and 0). This is exactly what I did:

\intke^(-|x|)
k\inte^(-|x|)
=k[-e^(-|x|)]

k[-e^{(-|infinity|)} - (-e^(-|0|)] = 1
k[0 - (-1)] = 1
k[0 +1] = 1
k = 1

Please tell me exactly WHERE I went wrong and how to fix it. Also, can you tell me how to start the probability density of y=x²?

 

[Predz]

Anyone?

 

[LPerrott]

I told you exactly how to do it. When integrating a probability density function you need to integrate from -infty to infty, but since you have absolute value of x this is a problem, therefore you you just integrate 2 times the integral from 0 to infty. k=1/2 As far as the next part of the question I'm not sure how to do it. It doesn't really make sense to me. Are you sure it isn't asking Y<=X^2 or Y>=X^2 ?

 

[LCKurtz]

I re-did it again (this time using infinity and 0). This is exactly what I did:

\intke^(-|x|)
k\inte^(-|x|)
=k[-e^(-|x|)]

k[-e^{(-|infinity|)} - (-e^(-|0|)] = 1
k[0 - (-1)] = 1
k[0 +1] = 1
k = 1

Please tell me exactly WHERE I went wrong and how to fix it. Also, can you tell me how to start the probability density of y=x²?

Where you went wrong is that the antiderivative of e^-|x| is not - e^-|x| on (-∞,∞). You have to look at two cases, x < 0 and x > 0.

[Edit] On (0,∞) just use |x| = x so the above is OK. My comments apply to x < 0 in particular. Do it by replacing |x| with its value.