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Suppose $X_1, X_2, \dots, X_n$ are i.i.d r.v.s with exponential distribution $E(\lambda) $. Let $S_n = X_1 + \dots + X_n$. Compute:

  1. $\mathbb{E}(S_n|X_1)$
  2. $\mathbb{E}(S_n^2|X_1)$
  3. $\mathbb{E}(S_n|S_k)$ where $n \geq k$

I think I got the first part: $$ \mathbb{E}(S_n|X_1) = \mathbb{E}(X_1 + \dots + X_n|X_1) = \mathbb{E}(X_1|X_1) + \mathbb{E}(X_2|X_1) + \dots + \mathbb{E}(X_n|X_1) $$ Now, $\mathbb{E}(X_1|X_1)$ is just $X_1$. The rest $n-1$ components will have the same value, namely if X, Y are i.i.d.:

$$ \mathbb{E}(X|Y) = \int_\mathbb{R} x g_{x|Y}dx = \int_0^{\infty} x \frac{\lambda^2e^{-\lambda (x + y)}}{\lambda e^{-\lambda y}} dx = \int_0^{\infty} x \lambda e^{-\lambda x }dx = \mathbb{E}X = \frac{1}{\lambda} $$

So finally $$ \mathbb{E}(S_n|X_1) = X_1 + \frac{n-1}{\lambda} $$

Is that a correct answer? How can I proceed with the next parts?

blahblah
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2 Answers2

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Your answer is correct.

2) $$ \mathbb E[S_n^2 | X_1] = \mathbb E[(S_n-X_1 + X_1)^2 | X_1] = \mathbb E[(S_n-X_1)^2 | X_1] + 2\mathbb E[X_1(S_n-X_1)|X_1] +\mathbb E[X_1^2|X_1] = \mathbb E[(\sum_{j=2}^n X_j)^2)]+ 2X_1\mathbb E[(S_n-X_1)|X_1] + X_1^2 = \sum_{2\leq i,j\leq n } \mathbb E[X_iX_j] +2X_1\frac{n-1}{\lambda} + X_1^2 = (n-1) \mathbb E[X_2^2] +\frac{(n-1)(n-2)}{\lambda^2} + 2X_1\frac{n-1}{\lambda} + X_1^2 $$ where $\mathbb E X_2^2 = \mathbb D^2(X_2) + \mathbb E^2X_2 = \frac{2}{\lambda^2}$.

3) $$ \mathbb E[S_n|S_k] = \mathbb E[(S_n-S_k)+ S_k | S_k] = \frac{n-k}{\lambda} + S_k $$

Mick
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  • Could you show how did it work here: $ \sum_{2\leq i,j\leq n } \mathbb E[X_iX_j]= (n-1) \mathbb E[X_2^2] +\frac{(n-1)(n-2)}{\lambda^2}$? I assume this has something to do with https://math.stackexchange.com/questions/1187687/expansion-of-the-square-of-the-sum-of-n-numbers but I cant see it right now. – blahblah Apr 19 '20 at 12:27
  • seperate the sum into two sum: when $i=j$ and when $i\neq j$. For $i=$ use identical distribution, for $i\neq j$ use the independence property. – Mick Apr 19 '20 at 13:31
  • $\mathbb E X_2^2 = \mathbb D^2(X_2) + \mathbb E^2X_2 = \frac{1+\lambda}{\lambda^2}$. But $Var(X_2)=\frac{1}{\lambda^2}$ and $ \mathbb E^2 (X_2) = \frac{1}{\lambda^2}$ which would make it sum to $\frac{2}{\lambda^2}$, wouldn't it? – blahblah Apr 19 '20 at 15:13
  • yes, that's a typo, I corrected it. – Mick Apr 19 '20 at 17:22
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$$\mathbb{E}\left[S_{n}\mid S_{k}=x\right]=\mathbb{E}\left[x+X_{k+1}+\cdots+X_{n}\mid S_{k}=x\right]=\mathbb{E}\left(x+X_{k+1}+\cdots+X_{n}\right)=x+\frac{n-k}{\lambda}$$ and: $$\mathbb{E}\left[S_{n}^{2}\mid S_{k}=x\right]=\mathbb{E}\left[\left(x+X_{k+1}+\cdots+X_{n}\right)^{2}\mid S_{k}=x\right]=\mathbb{E}\left(x+X_{k+1}+\cdots+X_{n}\right)^{2}=$$$$\mathsf{Var}\left(x+X_{k+1}+\cdots+X_{n}\right)+\left[\mathbb{E}\left(x+X_{k+1}+\cdots+X_{n}\right)\right]^{2}=\frac{n-k}{\lambda^{2}}+\left(x+\frac{n-k}{\lambda}\right)^{2}$$

From this we conclude that:

  • $\mathbb{E}\left[S_{n}\mid S_{k}\right]=S_{k}+\frac{n-k}{\lambda}$
  • $\mathbb{E}\left[S_{n}^{2}\mid S_{k}\right]=\frac{n-k}{\lambda^{2}}+\left(S_{k}+\frac{n-k}{\lambda}\right)^{2}$

To find $\mathbb{E}\left[S_{n}\mid X_{1}\right]$ and $\mathbb{E}\left[S_{n}^{2}\mid X_{1}\right]$ substitute $k=1$ and $S_{1}=X_{1}$.

drhab
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  • Thank you for the contribution. I also need to see what is $\mathbb{E}[e^{-S_n}|S_k]$. Would you be kind enough to derive that too?

    My attempt: $\mathbb{E}[e^{-S_n}|S_k] = \mathbb{E}[e^{-S_n+S_k-S_k}|S_k]=\mathbb{E}[e^{-S_n+S_k}e^{-S_k}|S_k]=\mathbb{E}[e^{-X_{k+1}-\dots-X_n}|S_k]e^{-S_k}=e^{-X_{k+1}-\dots-X_n}e^{-S_k} = e^{-S_n}$ seems wrong to me.

     – blahblah Apr 21 '20 at 07:43
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    As in the answers above begin with working out of: $$\mathbb{E}\left[h\left(S_{n}\right)\mid S_{k}=x\right]=\mathbb{E}\left[h\left(S_{k}+S_{n}-S_{k}\right)\mid S_{k}=x\right]=\mathbb{E}\left[h\left(x+S_{n}-S_{k}\right)\mid S_{k}=x\right]=$$$$\mathbb{E}h\left(x+S_{n}-S_{k}\right)$$In the case where $h$ is prescribed by $y\mapsto e^{-y}$ this will lead (by using independence and symmetry) to $$\mathbb{E}\left[e^{-S_{n}}\mid S_{k}=x\right]=e^{-x}\left(\mathbb{E}e^{-X_{1}}\right)^{n-k}=e^{-x}\left(\frac{\lambda}{1+\lambda}\right)^{n-k}$$ – drhab Apr 21 '20 at 11:52
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    Then conclude that: $$\mathbb{E}\left[e^{-S_{n}}\mid S_{k}\right]=e^{-S_{k}}\left(\frac{\lambda}{1+\lambda}\right)^{n-k}$$ – drhab Apr 21 '20 at 11:53