问题
I came across the following question:
Using rand() function, generate a number with expected value k. Options are:
1)
int GetRandom(int k)
{
v=0;
while(rand()<1.0f/(float)k)
v++;
return v;
}
2)
int GetRandom(int k)
{
v=0;
while(rand()<(1-1.0f/(float)k))
v++;
return v;
}
3)
int GetRandom(int k)
{
v=0;
while(rand() > (1-1.0f/(float)(k+1)))
v++;
return v;
}
1) seemed like the correct answer. Examining the outcome for specific values of k
seems to indicate this is the not the case. (I set k=3
. The frequency distribution of values for 100000
trials can be seen in the image below

How would one do this ?
The question is somewhat similar to this one.
回答1:
You want (2). This samples a Geometric Distribution (link) with mean k
.
A geometric distribution represents an experiment of this kind:
- A certain event happens repeatedly, with an outcome that is either 0 or 1
- The outcome of an event is 1 with probability
p
and 0 with probability1-p
- What is the index of the first event with an outcome of 1?
So if X ~ G(p)
, where X
is a random variable and p
is the probability above, then X
represents "What is the index of the first event with an outcome of 1?" The expectation is E[X] = 1/p
.
Given this information it should now be clear that the following represents a sampling of the random variable X
with p = 1/k
(and is equivalent to (2)).
int Sample(int k)
{
int v = 1;
while (true)
{
//outcome is true with probability p = 1/k
bool outcome = rand() < 1 / (double)k;
if (outcome)
return v;
else
v++;
}
}
Be aware that looking at the peak (mode) and expectation of the distribution are not the same thing. The peak of the geometric distribution is always going to be at 1!
来源:https://stackoverflow.com/questions/17223482/using-random-number-generator-to-generate-a-number-with-a-given-expected-value