Repeatedly multiplying by numbers taken from a uniform distribution
Motivation: In a software project I know well, for testing purposes, the evolution of a particular price is simulated by repeatedly multiplying by a random number picked between $0.995$ and $1.005$.
Let us formalize this.
Let $(X_i)_{i=1}^infty$ be a sequence of independent random variables, each with a (continuous) uniform distribution in $[0.995,1.005]$,
$$X_i sim U([0.995,1.005])$$
For each $ninmathbb{N}$ we define the random variable
$$P_n = prod_{i=1}^n X_i$$
which represents the price at day $n$.
I call for a proof of the fact that the price tends to zero as $n$ grows without bound. One way of formalizing this is to define a map $P_infty$ by:
$$P_infty : omegamapstobegin{cases}
lim_{ntoinfty}P_n(omega),&text{if the limit exists} \
text{undefined},&text{otherwise}
end{cases}$$
where $omega$ is an abstract outcome of the probability space. Then can you show that $P_infty$ is defined almost everywhere (and is measurable) so that we can see $P_infty$ as a random variable, and prove that the distribution of $P_infty$ is that of a constant, i.e.
$$mathrm{Pr}(P_infty = 0) = 1$$
where $mathrm{Pr}$ is the probability measure?
Bonus question: One way to "fix" the issue with the price going towards zero, would be to change the distribution of the $X_i$ into $X_i=exp(D_i)$ where the $D_i$ are independent and all distributed $D_i sim U([-0.005,0.005])$.
However, suppose for the purpose of the exercise, that we were required to keep the $X_i$ uniformly distributed, and could only change the lower endpoint, into, say, $alpha$. So now
$$X_i sim U([alpha,1.005])$$
and otherwise everything is as before. For too low values of $alpha$, $P_infty$ will be $0$ as before. But for higher values of $alpha$, it appears that $P_n$ would (almost surely) go to $+infty$ instead.
So the bonus question is, what "Goldilocks" value of $alpha$ should be chosen so that $P_infty$ is neither $0$, nor $+infty$?
probability-theory probability-distributions random-walk
asked Dec 8 at 10:34
Jeppe Stig Nielsen
3,2281123
add a comment |
Motivation: In a software project I know well, for testing purposes, the evolution of a particular price is simulated by repeatedly multiplying by a random number picked between $0.995$ and $1.005$.
Let us formalize this.
Let $(X_i)_{i=1}^infty$ be a sequence of independent random variables, each with a (continuous) uniform distribution in $[0.995,1.005]$,
$$X_i sim U([0.995,1.005])$$
For each $ninmathbb{N}$ we define the random variable
$$P_n = prod_{i=1}^n X_i$$
which represents the price at day $n$.
I call for a proof of the fact that the price tends to zero as $n$ grows without bound. One way of formalizing this is to define a map $P_infty$ by:
$$P_infty : omegamapstobegin{cases}
lim_{ntoinfty}P_n(omega),&text{if the limit exists} \
text{undefined},&text{otherwise}
end{cases}$$
where $omega$ is an abstract outcome of the probability space. Then can you show that $P_infty$ is defined almost everywhere (and is measurable) so that we can see $P_infty$ as a random variable, and prove that the distribution of $P_infty$ is that of a constant, i.e.
$$mathrm{Pr}(P_infty = 0) = 1$$
where $mathrm{Pr}$ is the probability measure?
Bonus question: One way to "fix" the issue with the price going towards zero, would be to change the distribution of the $X_i$ into $X_i=exp(D_i)$ where the $D_i$ are independent and all distributed $D_i sim U([-0.005,0.005])$.
However, suppose for the purpose of the exercise, that we were required to keep the $X_i$ uniformly distributed, and could only change the lower endpoint, into, say, $alpha$. So now
$$X_i sim U([alpha,1.005])$$
and otherwise everything is as before. For too low values of $alpha$, $P_infty$ will be $0$ as before. But for higher values of $alpha$, it appears that $P_n$ would (almost surely) go to $+infty$ instead.
So the bonus question is, what "Goldilocks" value of $alpha$ should be chosen so that $P_infty$ is neither $0$, nor $+infty$?
probability-theory probability-distributions random-walk
asked Dec 8 at 10:34
Jeppe Stig Nielsen
3,2281123
1
Use the law of large numbers for $log P_n$.
– zhoraster
Dec 8 at 13:11
@zhoraster That sounds right! If $X_i sim U([alpha, 1.005])$, is it true that we have $$E(log X_i) = int_alpha^{1.005} frac{log x}{1.005 - alpha} dx?$$ We know an antiderivative for $log x$ and can plug in. Then set expression equal to zero (we want the expectation of $log X_i$ to be zero) and solve for $alpha$. From that I get (using numerical method): $$alpha = 0.995008319483677112ldots$$ Could that be correct?
– Jeppe Stig Nielsen
Dec 8 at 14:56
It does seem correct to me.
– zhoraster
Dec 8 at 19:30
add a comment |
Motivation: In a software project I know well, for testing purposes, the evolution of a particular price is simulated by repeatedly multiplying by a random number picked between $0.995$ and $1.005$.
Let us formalize this.
Let $(X_i)_{i=1}^infty$ be a sequence of independent random variables, each with a (continuous) uniform distribution in $[0.995,1.005]$,
$$X_i sim U([0.995,1.005])$$
For each $ninmathbb{N}$ we define the random variable
$$P_n = prod_{i=1}^n X_i$$
which represents the price at day $n$.
I call for a proof of the fact that the price tends to zero as $n$ grows without bound. One way of formalizing this is to define a map $P_infty$ by:
$$P_infty : omegamapstobegin{cases}
lim_{ntoinfty}P_n(omega),&text{if the limit exists} \
text{undefined},&text{otherwise}
end{cases}$$
where $omega$ is an abstract outcome of the probability space. Then can you show that $P_infty$ is defined almost everywhere (and is measurable) so that we can see $P_infty$ as a random variable, and prove that the distribution of $P_infty$ is that of a constant, i.e.
$$mathrm{Pr}(P_infty = 0) = 1$$
where $mathrm{Pr}$ is the probability measure?
Bonus question: One way to "fix" the issue with the price going towards zero, would be to change the distribution of the $X_i$ into $X_i=exp(D_i)$ where the $D_i$ are independent and all distributed $D_i sim U([-0.005,0.005])$.
However, suppose for the purpose of the exercise, that we were required to keep the $X_i$ uniformly distributed, and could only change the lower endpoint, into, say, $alpha$. So now
$$X_i sim U([alpha,1.005])$$
and otherwise everything is as before. For too low values of $alpha$, $P_infty$ will be $0$ as before. But for higher values of $alpha$, it appears that $P_n$ would (almost surely) go to $+infty$ instead.
So the bonus question is, what "Goldilocks" value of $alpha$ should be chosen so that $P_infty$ is neither $0$, nor $+infty$?
probability-theory probability-distributions random-walk
asked Dec 8 at 10:34
Jeppe Stig Nielsen
3,2281123
Motivation: In a software project I know well, for testing purposes, the evolution of a particular price is simulated by repeatedly multiplying by a random number picked between $0.995$ and $1.005$.
Let us formalize this.
Let $(X_i)_{i=1}^infty$ be a sequence of independent random variables, each with a (continuous) uniform distribution in $[0.995,1.005]$,
$$X_i sim U([0.995,1.005])$$
For each $ninmathbb{N}$ we define the random variable
$$P_n = prod_{i=1}^n X_i$$
which represents the price at day $n$.
I call for a proof of the fact that the price tends to zero as $n$ grows without bound. One way of formalizing this is to define a map $P_infty$ by:
$$P_infty : omegamapstobegin{cases}
lim_{ntoinfty}P_n(omega),&text{if the limit exists} \
text{undefined},&text{otherwise}
end{cases}$$
where $omega$ is an abstract outcome of the probability space. Then can you show that $P_infty$ is defined almost everywhere (and is measurable) so that we can see $P_infty$ as a random variable, and prove that the distribution of $P_infty$ is that of a constant, i.e.
$$mathrm{Pr}(P_infty = 0) = 1$$
where $mathrm{Pr}$ is the probability measure?
Bonus question: One way to "fix" the issue with the price going towards zero, would be to change the distribution of the $X_i$ into $X_i=exp(D_i)$ where the $D_i$ are independent and all distributed $D_i sim U([-0.005,0.005])$.
However, suppose for the purpose of the exercise, that we were required to keep the $X_i$ uniformly distributed, and could only change the lower endpoint, into, say, $alpha$. So now
$$X_i sim U([alpha,1.005])$$
and otherwise everything is as before. For too low values of $alpha$, $P_infty$ will be $0$ as before. But for higher values of $alpha$, it appears that $P_n$ would (almost surely) go to $+infty$ instead.
So the bonus question is, what "Goldilocks" value of $alpha$ should be chosen so that $P_infty$ is neither $0$, nor $+infty$?
probability-theory probability-distributions random-walk
probability-theory probability-distributions random-walk
asked Dec 8 at 10:34
Jeppe Stig Nielsen
3,2281123
asked Dec 8 at 10:34
Jeppe Stig Nielsen
3,2281123
asked Dec 8 at 10:34
Jeppe Stig Nielsen
3,2281123
asked Dec 8 at 10:34
Jeppe Stig Nielsen
3,2281123
asked Dec 8 at 10:34
Jeppe Stig Nielsen
3,2281123
3,2281123
1
Use the law of large numbers for $log P_n$.
– zhoraster
Dec 8 at 13:11
@zhoraster That sounds right! If $X_i sim U([alpha, 1.005])$, is it true that we have $$E(log X_i) = int_alpha^{1.005} frac{log x}{1.005 - alpha} dx?$$ We know an antiderivative for $log x$ and can plug in. Then set expression equal to zero (we want the expectation of $log X_i$ to be zero) and solve for $alpha$. From that I get (using numerical method): $$alpha = 0.995008319483677112ldots$$ Could that be correct?
– Jeppe Stig Nielsen
Dec 8 at 14:56
It does seem correct to me.
– zhoraster
Dec 8 at 19:30
add a comment |
1
Use the law of large numbers for $log P_n$.
– zhoraster
Dec 8 at 13:11
@zhoraster That sounds right! If $X_i sim U([alpha, 1.005])$, is it true that we have $$E(log X_i) = int_alpha^{1.005} frac{log x}{1.005 - alpha} dx?$$ We know an antiderivative for $log x$ and can plug in. Then set expression equal to zero (we want the expectation of $log X_i$ to be zero) and solve for $alpha$. From that I get (using numerical method): $$alpha = 0.995008319483677112ldots$$ Could that be correct?
– Jeppe Stig Nielsen
Dec 8 at 14:56
It does seem correct to me.
– zhoraster
Dec 8 at 19:30
1
1
Use the law of large numbers for $log P_n$.
– zhoraster
Dec 8 at 13:11
Use the law of large numbers for $log P_n$.
– zhoraster
Dec 8 at 13:11
@zhoraster That sounds right! If $X_i sim U([alpha, 1.005])$, is it true that we have $$E(log X_i) = int_alpha^{1.005} frac{log x}{1.005 - alpha} dx?$$ We know an antiderivative for $log x$ and can plug in. Then set expression equal to zero (we want the expectation of $log X_i$ to be zero) and solve for $alpha$. From that I get (using numerical method): $$alpha = 0.995008319483677112ldots$$ Could that be correct?
– Jeppe Stig Nielsen
Dec 8 at 14:56
@zhoraster That sounds right! If $X_i sim U([alpha, 1.005])$, is it true that we have $$E(log X_i) = int_alpha^{1.005} frac{log x}{1.005 - alpha} dx?$$ We know an antiderivative for $log x$ and can plug in. Then set expression equal to zero (we want the expectation of $log X_i$ to be zero) and solve for $alpha$. From that I get (using numerical method): $$alpha = 0.995008319483677112ldots$$ Could that be correct?
– Jeppe Stig Nielsen
Dec 8 at 14:56
It does seem correct to me.
– zhoraster
Dec 8 at 19:30
It does seem correct to me.
– zhoraster
Dec 8 at 19:30
add a comment |
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1
Use the law of large numbers for $log P_n$.
– zhoraster
Dec 8 at 13:11
@zhoraster That sounds right! If $X_i sim U([alpha, 1.005])$, is it true that we have $$E(log X_i) = int_alpha^{1.005} frac{log x}{1.005 - alpha} dx?$$ We know an antiderivative for $log x$ and can plug in. Then set expression equal to zero (we want the expectation of $log X_i$ to be zero) and solve for $alpha$. From that I get (using numerical method): $$alpha = 0.995008319483677112ldots$$ Could that be correct?
– Jeppe Stig Nielsen
Dec 8 at 14:56
It does seem correct to me.
– zhoraster
Dec 8 at 19:30