为什么答案中E(exp(epsilont+h))=sigma^2/2

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？

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

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！

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？