last day (14 days later) » 

10:23
1
A: If $X,Y$ are geometric random variables, and $Z = X+Y$, then what is $P\{X= x, Z=z\}$?

probablyme1) At the moment you just have probability statement. You should provide an actual probability (plug into the pmf). 2) You want to recall that if $X,Y$ are iid, then the sum follows a negative binomial distribution, $$X+Y\sim \text{NB}(r = 2, p).$$ Further, we have that, assuming the support for...

The bottom 2 steps confuse me. X , Y has geometric distribution, I thought the formula for the mass function is (1-p)^k-1*p? How come your answer is (1-p)^k*p? Also, lots of textbooks use another formula for negative binomial distribution, when I use that formula, the answer is 1/(n-1). I am so confused. Please help!
If the possible values of $X$ and $Y$ are $\{0,1,2,3,\dotsc\}$, then $P(X = k) = (1-p)^{k} p$. If the possible values are $\{1,2,3,\dotsc\}$, then $P(X = k) = (1-p)^{k-1}p$. Instead of memorizing formulas, you should try to understand what they mean. In the first case, $X$ counts the number of failures until the first success; in the second, it counts the number of trials.
The same can be said about the negative binomial distribution.
Thanks, love you
Under the arrow that is under the $0$ to the left of my response, there is a check mark. You can show love by clicking it. It will then turn green and I will feel all warm and fuzzy inside if you do :)
Also, I am a bear (my school animal) :D
Are you telling me you and me are perfect match? Are you telling me you are from UCB?
WHERE AM I!
WHERE AM I
10:24
The chat.
WHERE THE BEEP BEEP BEEP AM I
In the world? I don't know.
hohohohoh
great
sweet
awesome
You are silly. haha. I am currently an undergrad at UCB. You?
shut the beep beep beep and let me kiss you all over!
you saved me
awesome
10:27
You are really silly. I doubt I saved you. What time is it at your place? It is 3 am here.
but ! i still have lots of question to finish. the deadline is monday
....................3am
doing what>
?
I doubt mind helping you. I am bored, not sleepy yet. I am lurking MSE to pass the time.
ok
let me ask you a question first
why the beep are you still awake?
what is the workload like in UCB
i am in new zealand
third year undergra
if you just left, i am gonna cry
I woke up really late yesterday. I overslept, so I'm not tired yet. It's really hard, there is so much work. So much work, that's all I can say. Cool! never met anyone from NZ before.
On the right side, the little avatars say who is in chat.
who?
who is in chat?
10:33
You and me.
you and me?
great!
and
are you in your fouth year
and it's 11pm sat
10:34
Wow!
so it is the last semester?
Yes. And it is 3am Saturday here.
so i need some help
Twelve independent random variables, each uniformly distributed over the interval
(0, 1], are added, and 6 is subtracted from the total. Determine the mean and variance
of the resulting random variable.
is the mean 0?
is the variance just 0.5
oh no
10:38
6 is just the number 6 right?
that is what i am trying to figure out
is 6 is the number
then mean is 0
the variance is 1
I got mean 0.
ok, let's assume 6 is just the number 6
yeah, and i got variance 1.
great!
you are such a perfect match
i like bear
10:42
At the moment, yes.
i like cute little animal
Me too.
i have a doggie
lucky!!
10:43
What kind? I don't.
I want one though
:(
west highland terrier
wow, cool. I bet it's cute. I want a corgi! They are so funny.
are you gonna do phd or not?
anyway, they pay you so you can leave off the campus and have a doggie
what did you just say !!!!!!!!!!!!!!!!!!!!!!!!
Yes, I want to, but not soon. I want to take a break from school for a couple of years. Really????????????????????
shut the beep! i wanted a corgi!!!!
you and me are perfect match
awesome
10:45
Hahaha, stop saying that. You are silly.
really what? pay you ? of course they pay you to do phd. don't you know that
ok, back to a serious topic
what is the distribution of the sum of n Poisson distribution
Are they independent? You should know this.
yes, i have the question here
The number of claims Z lodged in insurance company on each working day is Poisson
random variable with parameter λ:
P(Z = k) = λ
k
k!
e
−λ
, k = 0, 1, 2, . . . .
What is the distribution of the number of claims lodged during n working days? Prove
your answer.
i thought the answer is just the sum of lambas, but is that this easy?
is it a trap
Well, if I am reading the question correctly, then yes you are correct. It does not seem to be a trap.
In other words, it should be a poisson distribution with rate 7\lambda
what does rate 7/lambda mean?
10:54
I don't think chat supports formatting, it should be $7\lambda$

or just 7λ
oh, sorry, there are n working days, so it should be

do i use mgf to prove it?
Um, I don't think so.
................just did
any suggestion?
Oh, ok great. None
...............hahah
why none
hahaha
11:03
I had no suggestion.
:D it is ok
just did mgf, that will be fine
are you living in the hall
Ok, sounds good actually.

No, I rent out a room in a house.
Do you want to be pen pals? I might be going to sleep soon, but you seem friendly.
i would love to!
how do we contact each other
don't tell me you are chinese!
11:08
no
hahaha
do you have msn
facebook
nope and nope
but i like chating with you!
:(
hahaha, well, you have my email, I'm sure I can get msn or something. But I refuse to have a facebook, hahaha
promise?
ok, before you go to bed, check this out
11:11
ok
that will keep you smile during sleep
hahahahahaha, "my butt feels weird!"
and check out meow version
it is recommended next to it
ok
hahaha, wow
good night
11:16
ok, send me an email. good night, hope to help you soon!
let me try
just did
hahaah
ok, I got it. maybe I can help you tomorrow. Goodnight.
great
see you

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