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SkipperLin · 2020年03月05日

问一道题: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.

选项:

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[expϵT+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)]+σ2/2E_T[Y_{T+ h}] = exp[g(T + h)] +\sigma^2/2

And so:

ET[YT+h]=expET[lnYT+h]+σ2/2E_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.

请问一下 sigma 平方 / 2是怎么来的呀

1 个答案

袁园_品职助教 · 2020年03月06日

同学您好!

这道题的推导过程有一些疑问,我得到的结果是exp[(g(T+h)+σ^2/2)]

但是结论是一样的,就是当方差为零时两者相等。

关于推导过程中的疑问,我会和其他教研团队的成员讨论一下再给你答复,谢谢~加油!

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