Difference between $\mathbb{E}({T_1T_2\mid T)}$ and $\mathbb{E}({T_1^2\mid T)}$

Question

The problem is the following:

Let $(T_n)_{n>1}$ a succession of i.i.d.'s r.v with distribution $\exp(\lambda)$ and define $T=T_1+T_2+\cdots+T_n$

a) Calculate the conditional distribution of $T_1\mid T$

b) Calculate $\mathbb{E}(T_1\mid T)$

a) and b) are related and i got that for a) the distribution does not depend on the parameter $\lambda$ but only in the large of the sample and the distribution is a $Beta(1,n-1)$ times $T$**

But for the last two:

c) $\mathbb{E}({T_1T_2\mid T)}$

d) $\mathbb{E}({T_1^2\mid T)}$

Do they have to be equal? By i.i.d. I suppose both have to be equal but maybe there's a trick or something to obtain both or something like that.

Answer 1

$T_1$ and $T_2$ are independent, so the expectation of their product equals the product of their expectations. However, they are not conditionally independent.

Think about their joint conditional distribution some more....

$$\begin{align}f_{T_1,T_2\mid T}(x,y\mid z)=\dfrac{f_{T_1}(x)f_{T_2}(y)f_{T\mid T_1,T_2}(z\mid x,y)}{\int_0^{z-y}\int_0^{z-x-y} f_{T_1}(s)f_{T_2}(t)f_{T\mid T_1,T_2}(z\mid s,t)~\mathrm d t~\mathrm d s}\end{align}$$

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$T_1$ and $T_1$ are dependent (being surely equal), so the expectation of the square is not the square of the expectation.

Rather, since you have found the conditional distribution, use that : $$\mathsf E(T_1^2\mid T)=\mathsf{Var}(T_1\mid T)+\mathsf E^2(T_1\mid T)$$