问一道题：NO.PZ2020011101000019

问题如下：

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

选项：

解释：

A time trend model for $ln Y_t$ can be stated as:

$ln Y_t = g(t) + \epsilon_t, \epsilon ∼ N(0, \sigma^2)$ ,

where g(t) is a function of t.

So,

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

which gives

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

On the other hand:

$E_T[Y_{T+ h}] = E_T[exp(g(T + h) + \epsilon_{T+ h})] = exp(g(T + h) + E_T[exp \epsilon_{T+ h})]$ ,

which equals

$E_T[Y_{T+ h}] = exp[g(T + h)] +\sigma^2/2$

And so:

$E_T [Y_{T+ h}] = exp E_T[ln Y_{T+ h}] +\sigma^2/2$

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

请问这道题什么意思，可否讲解一下

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1 个答案

orange品职答疑助手 · 2020年02月20日

同学你好，这题是原版书上的题，有点超纲了，它问的是 exp( E ( lnY_T+H ) ) 和 E( Y_T+H))的大小关系。掌握一下结论就够了：E( Y_T+H))会大于等于exp( E ( lnY_T+H ) )，当方差为0时取等号

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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 7 · 回答

NO.PZ2020011101000019 老师，这道题的考点是啥？需要掌握哪个知识点呢？

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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！

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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！

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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？

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问一道题：NO.PZ2020011101000019

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