Conditional expectation in matlab

I have to calculate conditional expected value of a and b. a and b are random variables that are normally distributed with mean 1/2 and variance 1/2. The support for a and b, ie. Values of a and b, falls in the interval [0,1]. I have to calculate E(a|a>b) and E(b|a>b). How can I do this in Matlab?
Help @Torsten

6 件のコメント

Paul
Paul 2023 年 10 月 14 日
Hi Sabrina,
If a and b are normal, they each have support over the entire real line, so the statement about the interval of [0,1] is unclear.
I suspect some important information is missing from the question.
In any case, do you already have an approach that is supposed to be implemented in Matlab?
Sabrina Garland
Sabrina Garland 2023 年 10 月 14 日
@Paul Thanks for your reply. I don't want a and b to be on entire line. Can't I define it to be in the range [0,1]. I can't find code in Matlab which will help me define interval for normal distribution. All takes support to be from minus infinity to plus infinity. The variables a and b represent the gain percentage of two agent. Gain (profit) percentage cannot be negative. And the percentage value lies between 0 and 1.
William Rose
William Rose 2023 年 10 月 14 日
"I can't find code in Matlab which will help me define interval for normal distribution."
As @Paul correctly noted, you cannot find that because if the interval is not -If to +Inf, it would not be a normal distribution.
If you have an experimental situaiton where there is a ratio, and the ratio mut be between 0 and 1, then the ratio is not normally distributed. The classic example is the fraction of heads in N coin flips, which has a scaled binomial distribution.
William Rose
William Rose 2023 年 10 月 14 日
編集済み: William Rose 2023 年 10 月 14 日
[edit : fix typos]
another aspect of your original post which is impossible is
"... mean 1/2 and variance 1. The support for a and b, ie. Values of a and b, falls in the interval [0,1]."
Let's forget about whether the variables are normal or not. If the interval is [0,1], then it is impossible for the variance to be 1. Because var=1 means the average squared distance of each point from the mean is 1. For a variable with range [0,1], the mean will be between 0 and 1 (and you said mean=1/2, so that's fine). But the maximum distance from mean=1/2 to a point in this range is 1/2 (if the point is at 0 or 1). So the variance must be <1.
Sabrina Garland
Sabrina Garland 2023 年 10 月 14 日
Sorry! I mean variance of 1/2. Corrected it. Thanks for pointing it out @William Rose
William Rose
William Rose 2023 年 10 月 14 日
編集済み: William Rose 2023 年 10 月 14 日
@Sabrina Garland, you're welcome.
[edit: correct typo. I wrote var(1)=0.5 and I meant to write var(a)=0.5]
I think you meant that the standard deviation (not variance) is 1/2. Because if var(a)=0.5 then SD=1/sqrt(2)=0.707, which is still too much, if and range is [0,1].
If I am right that mean(a)=0.5 and s.d.(a)=0.5, and if range(a)=[0,1], then you must be talking about the Bernoulli distribution, i.e. the distribution of results from a single coin flip, for an evenly-weighted coin. The result of one flip is either 0 or 1, with mean 0.5 and variance=0.25, s.d.=0.5. Any other distribution with mean=0.5 will have an s.d. that is less than 0.5.
But you also said the random variable is a gain, so I don't know what to think.
Can you give a more detailed description of what you mean by the variable being the ratio of gains? How is each gain calculated? Whay can't the gain be more than 1 or less than 0. You also called it a profit. If you have an opportunity where the profit cannot be negative, there will be a lot of intereted people... :)

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