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The function
Another way is to look at the moment generating function of a stardard normal random variable,
which is
Say for example,
you had $100 in your bank accout 1 year ago
and the bank pays you 4% interest
now you have $104 in your account
you can say that you have 104% of the original amount you put in your bank 1 year ago
One of a very interesting property of Cauchy distribution is that the mode and median is 0.
But the mean is undefined.
Apparently intuition fails you here(!!) You have to look at the definition of expectation.
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the distribution you stated is called Cauchy distribution.
Wikipedia has a very good explanation for why the mean is undefined.
http://en.wikipedia.org/wiki/Cauchy_distribution
Read the section "Explanation of undefined moments"
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The second one is equal only when certain conditions are met because you are interchanging lim and integration sign, which is not always valid.
Here is a theorem I find in a book (The elements of integration by Bartle)
Suppose that for some t_0 in [a,b], the function x-> f(x, t_0) is integrable on R, that
exists on R x [a,b], and that there exists an integrable function g on R such thatThen
the answer is 32/3
What is your F(x)?
It should be x^3 + x
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It is because Z^2 is not closed under multiplication.
I have a feeling that we may be able to solve this without enumerating all the combination
Let y_max be the point of the largest of the six card
Let y be the point of a card drawn randomly from the deck
we can work out the probability mass function of y_max
say for example,
P(y_max = 10) = 6!/5! P(y = 10) (P(y <= 10 | a card with 10 points is drawn))^5
= 6*P(y = 10)
we can then work out the probability mass function of y_(5), which is the poit of the second largest card.
Haven't think about it carefully. But i think we might tackle this problem using the idea similar to order statistics.
Hi Bobbym.
I am new to this forum.
Getting bored with my exams and i found this when i surf the web aimlessly
You can imagine a simpler scenario.
There are 10 balls in an urn. 5 are red and 5 are white.
You get 1 point when you get a red and get 0 when you get a white.
You now draw two balls from the urn. What is the expected points that you will get?
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(2) is similar to (1).
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You can now try to work out the last step. (Bear in mind that X follows an exponential distribution)
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I used law of total probability and conditioned on the first toss.
Try to think about what the following probabilities are
P(A gets n heads before B gets m heads | first toss is head)
P(first toss is head)
you can do a first step analysis for this question.
P(A gets n heads before B gets m heads)
= P(A gets n before B gets m | first toss is head)P(first toss is head) + P(A gets n before B gets m | first toss is tail)P(first toss is tail)
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The idea of this question is that
the set
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