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Finding the cdf of a continuous variate

Let XX be a continuous random variate with pdf fX(x)=6x(1−x)f_X (x) = 6x(1-x) for 0≤x≤1 0\le x \le 1 and 0 0 otherwise.

I need to find the cdf FX(x)F_{X}(x) of XX.

I know that, in general, FX(x)=∫−∞xfX(t)dtF_{X}(x)= \displaystyle \int_{-\infty}^{x} f_{X}(t)dt

But in this case, I'm not sure where to integrate between. I tried integrating between [0, x] and [x, 1] but that just gives 1. So any help would be appreciated.
(edited 6 years ago)
Original post by MR1999
Let XX be a continuous random variate with pdf fX(x)=6x(1−x)f_X (x) = 6x(1-x) for 0≤x≤1 0\le x \le 1 and 0 0 otherwise.

I need to find the cdf FX(x)F_{X}(x) of XX.

I know that, in general, FX(x)=∫−∞xfX(t)dtF_{X}(x)= \displaystyle \int_{-\infty}^{x} f_{X}(t)dt

But in this case, I'm not sure where to integrate between. I tried integrating between [0, x] and [x, 1] but that just gives 1. So any help would be appreciated.


Your general formula is correct, and in this case the lower bound will be 0, since f(x) is 0 below x=0.

I can't make sense of "I tried integrating between [0, x] and [x, 1]" - sounds like you're integrating between two intervals???

Your limits should just be 0 and x, and you end up a function of x.

I'm going to have to keep my responses brief, as I keep getting "Service not available", and even when it is, the screen and cursor are jumping around all over as it's so f**king slow!
(edited 6 years ago)
Original post by MR1999
Let XX be a continuous random variate with pdf fX(x)=6x(1−x)f_X (x) = 6x(1-x) for 0≤x≤1 0\le x \le 1 and 0 0 otherwise.

I need to find the cdf FX(x)F_{X}(x) of XX.

I know that, in general, FX(x)=∫−∞xfX(t)dtF_{X}(x)= \displaystyle \int_{-\infty}^{x} f_{X}(t)dt

But in this case, I'm not sure where to integrate between. I tried integrating between [0, x] and [x, 1] but that just gives 1. So any help would be appreciated.


Just evaluate ∫0x6t(1−t).dt\displaystyle \int_0^x 6t(1-t).dt, which is valid for x∈[0,1] x \in [0,1]

EDIT: Obviously integrating over [0,x] then adding on the integral over [x,1] will give you 1. This is the same as integrating your density function over R\mathbb{R} in this case, and we know this must be 1 since you're summing all the probabilities!
(edited 6 years ago)

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