so from that i created the PMF, not pdf since this is a discrete distribution
i just don't know how im suppose to get the cdf from the pmf i had posted... hints?
for the cdf should I draw a graph of the pmf and then integrate somehow?
its so much easier when its converting a pdf to cdf, im not remembering how to do it for a pmf..
help.
The cdf still has the same definition. It's just that instead of integrating from negative infinity up to a point 'x' you're summing up to the point 'x'. If you look at it from a measure theory standpoint it's essentially the same thing.
so for 0<=x<=1 it would be 0.01
and for 1<=x<=2 it would be 0.1 or 0.11? i was confused which numbers to add or subtract for the different probabilities in the pmf
ahh i had a feeling thats how it was suppose to be. because if i graph the pmf and draw a line diagonally where 2<=x<3 it would include the point x=0,1,2 underneath it, since the cdf is the area under the curve of the pmf.
thanks so much for the clarification
one more question
if this was instead of a discrete distribution but a uniform distribution on interval [0,2]
i got the pdf to be {0.25 if 0<=x<=2, 0 if otherwise}
now when I tried to get the cdf I had trouble with 2<=x<4
here is what I have so far for the cdf
{0 if x<=0, (1/8)x^2 if 0<=x<2, 1 if x>=4}
this is a joint probability so I made a 2x2 square for the pdf and tried to find the area of the triangle for each interval and that was how I got (1/8)x^2, but I am unsure how to get the cdf when 2<=x<4
any hints?
You are talking about the joint CDF or just the univariate case?this is a joint probability so I made a 2x2 square for the pdf and tried to find the area of the triangle for each interval
any thing I missed?
If you are talking about the univariate case, then
if this was instead of a discrete distribution but a uniform distribution on interval [0,2]
i got the pdf to be {0.25 if 0<=x<=2, 0 if otherwise}are incorrect.now when I tried to get the cdf I had trouble with 2<=x<4
here is what I have so far for the cdf
{0 if x<=0, (1/8)x^2 if 0<=x<2, 1 if x>=4}
im almost 100% sure i have the pdf right, FOR UNIFORM distribution on [0,2]
but any hints on the cdf for that? ive been trying to solve it visually with a graph of the interval which is a 2x2 square.
and i thought i was on a roll
sorry i was talking about joint independent variables since the system X=X1+X2
Im understanding this a lot better now. really appreciate the help
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