Conditional expected value of sum of exponential r.v.s

Question

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?

Answer 1

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

Answer 2

$$\mathbb{E}\left[S_{n}\mid S_{k}=x\right]=\mathbb{E}\left[x+X_{k+1}+\cdots+X_{n}\mid S_{k}=x\right]=\mathbb{E}\left(x+X_{k+1}+\cdots+X_{n}\right)=x+\frac{n-k}{\lambda}$$ 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 $S_{1}=X_{1}$.