Use the law of total expectation. E(T) = E(E(T|X>Y)) where the variable X>Y ranges over the values true and false with probability 1/2 for each. Each of the conditional distributions is symmetric about 5 so we have 5*.5 + 5*.5 = 5.
I think others below/above may have commented about using the "law of total probability". This is a compromise between there terse comments and the explanation in the video. Draw the line y = x to split the triangle (in the video) into two pieces. The max in the bottom triangle is x and the max in the top triangle in y. Now you can integrate to get the answer, rigorously with perhaps a shorter solution. But, you really do not need to integrate, do you? The integral in the bottom triangle is 1/2 the expected value of x in that triangle (1/2 because the area of the bottom triangle is 25 and the density function is 1/50). Observe that x is symmetric about 5, so half the expected value is 5/2. Similarly, the integral in the top triangle is half the expected value of y. Observe that y is symmetric about 5, and so, the integral is 5/2. Now the answer is 5/2 + 5/2 = 5. All this can be made mathematically rigorous.
These questions are so tricky!!! Because I recall another question where they say that the system fails if EITHER fails so definitelly be careful with the wording!!!
I have 2 questions: (1) shouldn't the piecewise F(t) integrated over the appropriate bounds and then summed, equal to 1? It adds up to 5 here. Just curious. (2) (at minute 15:30) When using the CDF to get the expected value, the formula in my book adds the value of lower bound to the integral. For the first piece, it was zero so it's okay, but shouldn't the second integral include a +5 before the integral sign? Thanks
Use the law of total expectation. E(T) = E(E(T|X>Y)) where the variable X>Y ranges over the values true and false with probability 1/2 for each. Each of the conditional distributions is symmetric about 5 so we have 5*.5 + 5*.5 = 5.
hey, just wanted to thank you for all these videos. I went through every practice solution
you made before taking exam P, and Passed!
Yunjin Jung were soa sample exams enough
Aarij Siddiqui i would say so and i also went through infinite actuary practice exams.
Maybe we can consider the center of mass of a triangle, x=5,y=5, and max(x,y)=5 because of uniform is equally likely
I think others below/above may have commented about using the "law of total probability". This is a compromise between there terse comments and the explanation in the video.
Draw the line y = x to split the triangle (in the video) into two pieces. The max in the bottom triangle is x and the max in the top triangle in y. Now you can integrate to get the answer, rigorously with perhaps a shorter solution.
But, you really do not need to integrate, do you? The integral in the bottom triangle is 1/2 the expected value of x in that triangle (1/2 because the area of the bottom triangle is 25 and the density function is 1/50). Observe that x is symmetric about 5, so half the expected value is 5/2.
Similarly, the integral in the top triangle is half the expected value of y. Observe that y is symmetric about 5, and so, the integral is 5/2. Now the answer is 5/2 + 5/2 = 5. All this can be made mathematically rigorous.
I used the same approach
i can only do this problem using lotp conditioning on y > x, x < y ea. w/1/2 probability
Harvinder Sandhu dame here I used LOTP
i believed i did this too, is it 2.5 for each triangle and you add them up to get 5?
"If X has a standard normal distribution and Y = e^x, what is the k-th moment of Y? I'm struggling with this question for some reason.
Thanks for your request, ill cover this ASAP!
Your requested example is up!
THANKS MAN, THIS QUESTION BUT I REALLY DONT GET HOW YOU CAME UP WITH THE LIMITS OF INTEGRATION FOR f(x)
These questions are so tricky!!! Because I recall another question where they say that the system fails if EITHER fails so definitelly be careful with the wording!!!
Indeed. Miss interpreting the question is an easy way to get the question wrong... I have done it many times.
I have 2 questions:
(1) shouldn't the piecewise F(t) integrated over the appropriate bounds and then summed, equal to 1? It adds up to 5 here. Just curious.
(2) (at minute 15:30) When using the CDF to get the expected value, the formula in my book adds the value of lower bound to the integral. For the first piece, it was zero so it's okay, but shouldn't the second integral include a +5 before the integral sign?
Thanks
Edina Dudas, the integral of the PDF over the domain should equal 1, not CDF. He is finding expected value when integrating the survival function.
The integral att the end only holds for values greater than 0!
time is always greater than 0
You took an analysis type of approach to the problem lol the little t is like epsilon
Honestly, my approach is probably over done for an exam setting but I really didn't like the solution given by soa.