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Finding the CDF

nacho

Active member
Sep 10, 2013
156
Please refer to the attached image.

For part a)

when I want to find the CDF, don't I simply take the indefinite integral of e^-|x|, multiply it by c and solve for that = 1?

I am unsure of how to take the integral for this, am i correct in saying it is -e^-x, for all x ?

that would leave me with c* -e^-x = 1, and c = 1/(-e^-x), wolfram says there are two separate results, but i am not sure why. and in that case, I would also have two separate results for c. how can that make sense?

for part b) i am unsure as to how to approach the question. could someone please guide me?

As always, your help is very much appreciated and invaluable.

Thanks
 

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Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Hello nacho,
for part a), that's not what you are meant to do. Remember that probability density functions must integrate to 1 over the entire domain, that is:

$$\int_{-\infty}^{+\infty} c e^{- |x|} ~ \mathrm{d} x = 1$$

Can you solve for $c$ now? (you don't need to know what a CDF is for this part of the question)

Next, the CDF is by nature a definite integral, since when you say "the CDF of $f_X$ is $F(x)$" then you are saying that:

$$F(x) = \int_{- \infty}^x f_X (t) ~ \mathrm{d} t$$

Note the variable $t$, it is not the same as $x$. Basically summing up the PDF from negative infinity to $x$, the CDF variable. So in your case:

$$F(x) = \int_{-\infty}^x c e^{- |t|} ~ \mathrm{dt}$$

Which will give you an expression for the CDF in terms of $x$, and then finding its mean and variance should be straightforward.

[HR][/HR]

For part b), I am not sure why you are asked to find the CDF first instead of the PDF, but there is probably some sort of theorem that relates the CDF of the transform of a random variable with the CDF of that random variable (which you found previously). Do your notes say anything?
 
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nacho

Active member
Sep 10, 2013
156
Hello nacho,
for part a), that's not what you are meant to do. Remember that probability density functions must integrate to 1 over the entire domain, that is:

$$\int_{-\infty}^{+\infty} c e^{- |x|} = 1$$

Can you solve for $c$ now? (you don't need to know what a CDF is for this part of the question)



[HR][/HR]

For part b), I am not sure why you are asked to find the CDF first instead of the PDF, but there is probably some sort of theorem that relates the CDF of the transform of a random variable with the CDF of that random variable (which you found previously). Do your notes say anything?
Thanks

What I wanted to clarify was, for part a)
this is what I did:

split up the integral to:

$$ C(int\e^x dx + \int e(^-x)dx) = 1 $$

which gives

$$ c = 1/{e^x-e^(-x)} $$
Is this correct?


god i'm bad at latex.

What i meant was c(int e^x dx + int e^-x dx) = 1
thus c = 1/(e^x -e^(-x)) ?


looking at my notes for part b) I can't seem to find any relationship.
 
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Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
No, it's not correct. You cannot find $c$ in terms of $x$, that doesn't make sense. Remember what I mentioned earlier: a PDF always has an area (under its curve) of $1$. That means that if you integrate your PDF from negative infinity to positive infinity and you get, say, $2c$, then you know that $2c = 1$ and so $c = \frac{1}{2}$. Basically the fact that the probability density function must integrate to $1$ puts conditions of what values $c$ can take (and, in fact, it can only take one value).

So it's a definite integral. Can you integrate $f(x) = c e^{-|x|}$ from negative infinity to positive infinity (find the area under this function's curve, in terms of $c$)? What do you get?

Rereading your post you are pretty close, but again you are not calculating a definite integral but only an indefinite one. You want to calculate:

$$\int_{-\infty}^{+\infty} c e^{-|x|} ~ \mathrm{d} x = 1$$

And you correctly identified that you should split it for $x < 0$ and $x > 0$. So:

$$\int_{-\infty}^{+\infty} c e^{-|x|} ~ \mathrm{d} x = \int_{-\infty}^{0} c e^{x} ~ \mathrm{d} x + \int_{0}^{+\infty} c e^{-x} ~ \mathrm{d} x = 1$$

I think this is what you were missing. Can you complete it now?

You are mixing up definite and indefinite integrals.
 
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nacho

Active member
Sep 10, 2013
156
ahhh,
thank you
that makes sense now, I see what you mean.

I was having trouble with the bounds when i split it up, so I thought it was just supposed to calculate the indefinite integral.

i understand now though!