开发者:上海品职教育科技有限公司 隐私政策详情

应用版本:4.2.11(IOS)|3.2.5(安卓)APP下载

Luhunlu · 2022年02月08日

老师,请问这道题到底是要掌握哪个知识点呀?

NO.PZ2020011101000019

问题如下:

When modeling lnYtln Y_t using a time trend model, what is the relationship between expET[lnYT+h]exp E_T[ln Y_{T+ h}] and ET[YT+h]E_T[Y_{T+ h}] for any forecasting period h? Are these ever the same? Assume the error terms is normally distributed around a mean of zero.

解释:

A time trend model for lnYtln Y_t can be stated as:

lnYt=g(t)+ϵt,ϵN(0,σ2)ln Y_t = g(t) + \epsilon_t, \epsilon ∼ N(0, \sigma^2),

where g(t) is a function of t.

So,

ET[lnYT+h]=g(T+h)E_T[ln Y_{T+ h}] = g(T + h),

which gives

expET[lnYT+h]=exp[g(T+h)]exp E_T[ln Y_{T+ h}] = exp [g(T + h)],

On the other hand:

ET[YT+h]=ET[exp(g(T+h)+ϵT+h)]=exp(g(T+h)+ET[exp epsilonT+h)]E_T[Y_{T+ h}] = E_T[exp(g(T + h) + \epsilon_{T+ h})] = exp(g(T + h) + E_T[exp \ epsilon_{T+ h})],

which equals

ET[YT+h]=exp[[g(T+h)+σ2/2]E_T\lbrack Y_{T+h}\rbrack=exp\left[\lbrack g(T+h)+\sigma^2/2\right]

And so:

ET[YT+h]=expET[[lnYT+h+σ2/2]E_T\lbrack Y_{T+h}\rbrack=expE_T\left[\lbrack lnY_{T+h}+\sigma^2/2\right]

These will be equal if the variance is zero (in other words, if the process is completely deterministic.

老师,这道题的考点是啥?需要掌握哪个知识点呢?

1 个答案

李坏_品职助教 · 2022年02月08日

嗨,努力学习的PZer你好:


这个题目考察的是对数正态分布的性质,稍微有点超纲了,是原版书的超纲题。


题目里的模型是:lnYt = g(t) + e,也就是yt服从对数正态分布。

根据上图,E(Yt)= exp(g(t) + 1/2 * σ^2)。 我们用t+h替换t就行了。也就是exp[(g(T+h)+σ^2/2)]

----------------------------------------------
就算太阳没有迎着我们而来,我们正在朝着它而去,加油!

  • 1

    回答
  • 1

    关注
  • 364

    浏览
相关问题

NO.PZ2020011101000019问题如下When moling lnYtln Y_tlnYt​ using a time trenmol, whis the relationship between expET[lnYT+h]exp E_T[ln Y_{T+ h}]expET​[lnYT+h​] anET[YT+h]E_T[Y_{T+ h}]ET​[YT+h​] for any forecasting perioh? Are these ever the same? Assume the error terms is normally stributearouna meof zero. A time trenmol for lnYtln Y_tlnYt​ cstateas: lnYt=g(t)+ϵt,ϵ∼N(0,σ2)ln Y_t = g(t) + \epsilon_t, \epsilon ∼ N(0, \sigma^2)lnYt​=g(t)+ϵt​,ϵ∼N(0,σ2),where g(t) is a function of t. So,ET[lnYT+h]=g(T+h)E_T[ln Y_{T+ h}] = g(T + h)ET​[lnYT+h​]=g(T+h), whigivesexpET[lnYT+h]=exp[g(T+h)]exp E_T[ln Y_{T+ h}] = exp [g(T + h)]expET​[lnYT+h​]=exp[g(T+h)],On the other hanET[YT+h]=ET[exp(g(T+h)+ϵT+h)]=exp(g(T+h)+ET[exp epsilonT+h)]E_T[Y_{T+ h}] = E_T[exp(g(T + h) + \epsilon_{T+ h})] = exp(g(T + h) + E_T[exp \ epsilon_{T+ h})]ET​[YT+h​]=ET​[exp(g(T+h)+ϵT+h​)]=exp(g(T+h)+ET​[exp epsilonT+h​)],whiequals ET[YT+h]=exp[[g(T+h)+σ2/2]E_T\lbraY_{T+h}\rbrack=exp\left[\lbrag(T+h)+\sigma^2/2\right]ET​[YT+h​]=exp[[g(T+h)+σ2/2]Anso: ET[YT+h]=expET[[lnYT+h+σ2/2]E_T\lbraY_{T+h}\rbrack=expE_T\left[\lbralnY_{T+h}+\sigma^2/2\right]ET​[YT+h​]=expET​[[lnYT+h​+σ2/2]These will equif the varianis zero (in other wor, if the process is completely terministi老师好,1、红线等式看不明白,是怎么等价过来的呢?2、蓝色线是为什么相等呢?能给个推导过程吗?

2024-08-13 16:28 6 · 回答

NO.PZ2020011101000019 E(exp(εt+h)))为什么不等于1而是σ^2呀?E(εt+h)不是等于0吗?谢谢!

2022-01-06 16:53 1 · 回答

NO.PZ2020011101000019 请问这步是怎么得到的?ET​[exp(g(T+h)+ϵT+h​)]=exp(g(T+h)+ET​[exp epsilonT+h​)] 以及后一项如何转化为下一步的sigma^2/2? 谢谢!

2021-09-23 22:20 2 · 回答

误差服从i,那误差的指数怎么求方差啊?能否讲一下sigma平方/2怎么来的呀?

2020-09-25 22:57 1 · 回答