Given: .5x, x E [0, 1] .5, x E [1, 2] .5(3-x), x E [2, 3] 0 otherwise We calculate the indefinite integrals of each, which I will mark with "I#", where # is the number, and I stands for integral: I1 = .25x^2 I2 = .5x I3 = 1.5x - .25x^2 Now evaluate each one within each interval. "E" stands for evaluated, "#" for number E1 = I1[0 to 1] (.25 - 0) = .25 E2 = I2[1 to 2] = (1 - .5) = .5 E3 = I3[2 to 3] = (2.25 - 2) = .25 In this process, we also found what it was evaluated at for each lower bound. Let's mark these as LI#: LI1 = 0, LI2 = .5, LI3 = 2 Resulting CDF is: 0, x < 0 I1, x E [0, 1] I2 - LI2 + E1, x E [1, 2] I3 - LI3 + (E1 + E2), x E [2, 3] 1, x > 3 Evaluated: 0 .25x^2 = x^2/4 .5x - .25 = x/2 - 1/4 1.5x - .25x^2 - 1.25 = 3/2x - x^2/4 - 5/4 1 SUMMARIZED: Current Integral (CI) - CI Lower Bound + Previous Integrals Evaluated
in continuous pdf function if you include or exclude the upper value of any interval in cdf does not matter though we usually follow one pattern like (a
The upper limit in CDF calculation is a variable because it represents the value up to which we are accumulating probability from the PDF. As the variable changes, the CDF captures the cumulative probability up to that point.
Thank you so much!!!! I've looking everywhere and only your video helped me to actually do the exercise. Your video is very much practical. Thanks!
we understand through examples...hence proved...thank you soo much maam
Great video mam…this made me remember my uni days again, miss your lectures ❤❤… keep going mam🫶🏽
Thanks a lot
Now I finally understand the concept behind it 😙😗😙😙
Glad to hear that
I was soo confused between those limits 😞 and thanks alot...for helping us thank you ma'am ❤
mam in last steps we have to add 1 in range 1 to 2, we consider 1 but not 2 in place of x. am i right or i miss something? of part 4
integration of function = 1/2 from 1 to 2 is equal to 1/2
@@maths.tutor4u1722 to x right instead of 2-3 bcz x lies b/w 2-3
@@ikramulhaq3124 yes and I rectified it in the next step while putting the limits in solution
I was very confesed why we are taking 1 at the last though it has 0 for x>3 but now i understood thank u mam
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Hare Krishna 🙏🏻🌸
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Wow super mam . Now I am very clear ❤
Awesome explanation 😊❤😊
Given:
.5x, x E [0, 1]
.5, x E [1, 2]
.5(3-x), x E [2, 3]
0 otherwise
We calculate the indefinite integrals of each, which I will mark with "I#", where # is the number, and I stands for integral:
I1 = .25x^2
I2 = .5x
I3 = 1.5x - .25x^2
Now evaluate each one within each interval. "E" stands for evaluated, "#" for number
E1 = I1[0 to 1] (.25 - 0) = .25
E2 = I2[1 to 2] = (1 - .5) = .5
E3 = I3[2 to 3] = (2.25 - 2) = .25
In this process, we also found what it was evaluated at for each lower bound. Let's mark these as LI#:
LI1 = 0, LI2 = .5, LI3 = 2
Resulting CDF is:
0, x < 0
I1, x E [0, 1]
I2 - LI2 + E1, x E [1, 2]
I3 - LI3 + (E1 + E2), x E [2, 3]
1, x > 3
Evaluated:
0
.25x^2 = x^2/4
.5x - .25 = x/2 - 1/4
1.5x - .25x^2 - 1.25 = 3/2x - x^2/4 - 5/4
1
SUMMARIZED: Current Integral (CI) - CI Lower Bound + Previous Integrals Evaluated
Can you explain how the inequalities can be inclusive on both ends for the cdf? Shouldn't some of them be just < instead of < or equal to? Thank you.
in continuous pdf function if you include or exclude the upper value of any interval in cdf does not matter though we usually follow one pattern like (a
Thank you so much ma'am. I had confusion with this which got sorted out.
Great Explanation ma'am👍🏼
Thanks
Mam can you tell me from which textbook this example was taken from ?
why in the cdf 4th integration the limit is 2 to x and not 2 to 3?
The upper limit in CDF calculation is a variable because it represents the value up to which we are accumulating probability from the PDF. As the variable changes, the CDF captures the cumulative probability up to that point.
Thank u so much! Good explanation :D
Awesome work ,,thanks maam
Very well explained. Thank you so much
Thanks a lot. U are so great ❤
supposed to be +2 not -2, for the last cdf
Thanks a lot mam
th-cam.com/video/qmzfbuTeCRw/w-d-xo.html my new video to find cdf using crisp and short method.....hope it will help you
Fantastic ❤
why in the 4th part we integrated from 2 to 3 and not۔
2 to x
its 2 to x actually by mistake written 3 but rectified in next step
@@maths.tutor4u172thank you mam
why in the 2nd part we add -1/2
why didn't we just add x2/4 +1/4
Thankyou so............. Much ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤
can see latest video uploaded with short cut method
Thank you!
Can you explain to me how we make integration of (x/2) to x^2/4
take 1/2 constant outside and integrate x dx , = 1/2 ( x^2/2)
@@maths.tutor4u172 got it thanks ❤️
Thank you so much!
🪐