How do you find the marginal distribution of a random variable, given a table of the joint probabilities of that random variable and another?
i.e. fY(y).
I see the solution but not sure where they are getting it from.
And we're using the convention X ~ exp(a) => f_{X}(x) = 1/a exp(-x/a)
But how do you find the conditions on y_{1} and y_{2} where this density is defined for?
50 % of the meals sold at Graylands Hospital contain rice.
Answer the following question, keeping in mind that your answer must be a number between 0 and 1, i.e do not use percentages. Round off the final result to 2dp, but don't round numbers during your calculations.
What is the probability that out of 78 meals sold at Graylands Hospital, 43 or more contain rice?
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Last edited by InteGrand; 20 Apr 2017 at 8:57 PM. Reason: Typo
U_{1}, ..., U_{n} are iid Unif(0,1)
So I followed their working up to
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Edit: Actually my bad on transcription error, their working stopped here
Last edited by leehuan; 21 Apr 2017 at 5:49 PM.
The integral you were thinking about is actually the "complementary CDF" or "survival function" (though the limits aren't right and I realised I misinterpreted your query before seeing your addendum), which is the probability Y_n > y. It's actually neater here find this survival function first and then just get the CDF of Y_n from that by taking its complement (subtracting it from 1). The reason it's nicer to work with the survival function here is that's we're dealing with a min function, and the minimum of some numbers is greater than something iff all the numbers are greater than that thing. If instead we were working with a max function, it'd be nicer to use the CDF, because a maximum of some numbers is less than or equal to something iff all the numbers are less than or equal to that thing.
Edit: I think I misinterpreted your query (I think I get it now after seeing your addendum). I addressed what I think was your query in my next post.
Last edited by InteGrand; 21 Apr 2017 at 6:06 PM.
I computed an MGF to be
Therefore it does not exist because it's not defined for any u in [interval containing 0] right?
Can't any quick way of doing it at the moment. Please let me know if any of this is wrong.
Since probabilities sum to 1, we have
By symmetry (i.e. Pascal's Triangle), we have
Then from (*), we have
Now, we compute the desired probability,
In retrospect, I should have wrote p = 0.5 = 1 - p and combined the product which would make it easier to read.
Last edited by He-Mann; 28 Apr 2017 at 9:13 PM.
Once again stuck on domains. I keep forgetting how linear transforms map regions so maybe just another 'by inspection' method please?
(Intuitive guess: -60 < z < 60. Just don't know how to get there)
It's quite easy with the linear transforms method. The sample space in the v-w space is just a square. A square gets mapped to a parallelogram in the u-z space (in general) by a linear transformation like the one you have. (Recall that in general, an invertible linear map from R^{n} to R^{n} will map lines to lines and parallelograms to parallelograms.)
To find the coordinates of the vertices this parallelogram, just find what points each of the vertices of the square in the v-w sample space get mapped to by the linear transformation. Then the interior of the sample space in the v-w space gets mapped to the inside of the parallelogram with those vertices in the u-z space.
Last edited by InteGrand; 22 Apr 2017 at 5:22 PM.
That's interesting, u - 60 < z \le u was the first thing I had.
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