Question

The distribution function of random variable X is given by F(X) = 1 - (1+X) e^-X >=0 then how to find density function, mean and variance of X ?

Answer 1

@prathameshtambe007 Hi :)
\(\huge \color{red}{\text{Welcome to Open Study}}\ddot\smile\)

Do you know that f(x) = F'(x) ?
where f(x) is density function and F(x) is distribution function.
so, you just need derivative of F(x), can you find ?

Answer 2

okay thats great I'll differentiate it. but then what about mean and variance?

Answer 3

mean is dsfined as
\(\huge m=\int_D xf(x)dx\)
where D is the domain of the function f(x)

Answer 4

and variance is
\(\huge V =\int_D x^2f(x)dx-m^2\)

Answer 5

yep this one I know but what should be the domain? is it from 0 to infinity? also what is f(x) a distribution function or a density function? I tried this thing with integrating distribution function with from 0 to infinity and I got the answer infinity which is not acceptable. What should I do?

Answer 6

f(x) = density function.
domain = 0 to infinity

Answer 7

what you got after differentiating ?

Answer 8

oh so its density function and I tried with distribution function. Just a moment I'll try is again and let you know

Answer 9

because i am getting a constant answer for mean.

Answer 10

take your time :)

Answer 11

I got X.e^-X as density function. is it correct?

Answer 12

absolutely! :)

Answer 13

oops then after multiplying it by X and integrating
m= integration 0 to infinity X^2. e^-X
= X^2 (- e^-X) - 2X (e^-X) + 2(-e^ -X) limits 0 to infinity
so if I put infinity at X I'll get 0 and if I punt 0 still I'll get 0 so the answer is zero. is it so

Answer 14

'if i put 0' <------try again
everything else is correct ^_^

Answer 15

oh yes got it the last term will be -2

Answer 16

so, m = 0- (-2) = 2 is the mean.

Answer 17

try variance now :)

Answer 18

variance is 6 ???

Answer 19

just one more look at the formula, what you got as 6 ?
and then, what will be the variance ?

Answer 20

shot forgot to substract m2 . varience will 6-4 = 2

Answer 21

you are \(\huge \color{red}{\checkmark}\)

Answer 22

good work!

Answer 23

Thanks a lot........ !!!!! @hartnn

Answer 24

welcome ....!!! @prathameshtambe007
oh, and since you are new here, if you have any doubts browsing this site, you can ask me, or to an ambassador (with an encircled A before their name).
Have fun learning with Open Study! :)
Hope you like this site.

Answer 25

(Y)

Answer 26

thanks