# Expected value of last Gamma RV in a sum

I've got a sum of $$X_i \sim \text{Gamma}(k, \theta)$$ i.i.d. random variables. I'm trying to find the expected value of the final $$X_i$$ that takes the sum above a certain value, i.e., to find the value of

$$E[X_n | X_1 + X_2 + \cdots + X_{n-1} < 1000, X_1 + X_2 + \cdots + X_n > 1000]$$ for integer $$n > 1$$.

For simplicity, denote $$X_n$$ by $$\xi$$ and $$X_1 + \cdots + X_{n - 1}$$ by $$\eta$$. By assumption and the summation property of Gamma distribution, we know that $$\xi \sim \Gamma(k, \theta)$$ and $$\eta \sim \Gamma((n - 1)k, \theta)$$. In addition, $$\xi$$ and $$\eta$$ are mutually independent. Thus the problem reduces to find the conditional expectation $$E[\xi | \eta < 1000, \xi + \eta > 1000]$$. Note that unlike the common case, this "conditional expectation" is conditioning on an event instead of a random variable ($$\sigma$$-algebra), thus the result is a scalar instead of a random variable.

First, by the expectation formula for non-negative random variables, \begin{align} & E[\xi | \eta < 1000, \eta + \xi > 1000] \\ = & \int_0^\infty P[\xi > t | \eta < 1000, \eta + \xi > 1000] dt \\ = & \frac{1}{P[\eta < 1000, \eta + \xi > 1000]}\int_0^\infty P[\xi > t, \eta < 1000, \eta + \xi > 1000] dt \tag{1} \end{align}

Edit: Just realized an analytical solution is possible by means of Gamma function and Beta function when the shape parameter $$k$$ is a positive integer. To save some typing, let's assume without generality that the scale parameter $$\theta \equiv 1$$, and denote $$1000$$ by $$\alpha$$. Moreover, let \begin{align} f_\xi(x) & = \frac{1}{\Gamma(k)}e^{-x}x^{k - 1} \equiv c_1 e^{-x}x^{k - 1} \quad (x > 0), \\ f_\eta(y) & = \frac{1}{\Gamma((n - 1)k)}e^{-y}y^{(n - 1)k - 1} \equiv c_2 e^{-y}y^{m - 1} \quad (y > 0) \\ \end{align} denote density functions of $$\xi$$ and $$\eta$$. Then $$P[\eta < \alpha, \xi + \eta > \alpha] = c_1c_2\int_0^\alpha\left[\int_{\alpha - y}^\infty e^{-x}x^{k - 1}dx\right] e^{-y}y^{m - 1}dy. \tag{2}$$

Make substitution $$t = x - (\alpha - y)$$ in the inner integral of $$(2)$$, it follows that \begin{align} & \int_{\alpha - y}^\infty e^{-x}x^{k - 1} dx \\ = & e^{y - \alpha} \int_0^\infty e^{-t}[t + (\alpha - y)]^{k - 1} dt \\ = & e^{y - \alpha} \sum_{i = 0}^{k - 1} \binom{k - 1}{i}(\alpha - y)^i\int_0^\infty e^{-t}t^{(k - i) - 1}dt \\ = & e^{y - \alpha} \sum_{i = 0}^{k - 1} \binom{k - 1}{i}(\alpha - y)^i\Gamma(k - i). \end{align} Therefore, the right hand side of $$(2)$$ equals to \begin{align} & c_1c_2 e^{-\alpha}\sum_{i = 0}^{k - 1}\binom{k - 1}{i}\Gamma(k - i)\int_0^\alpha y^{m - 1}(\alpha - y)^i dy \\ = & c_1c_2e^{-\alpha} \sum_{i = 0}^{k - 1}\binom{k - 1}{i}\Gamma(k - i)\alpha^{m - i}\int_0^1 z^{m - 1}(1 - z)^{i + 1 - 1} dz \\ = & c_1c_2e^{-\alpha} \sum_{i = 0}^{k - 1}\binom{k - 1}{i}\Gamma(k - i)\alpha^{m - i}B(m, i + 1) \\ = & c_1c_2e^{-\alpha} \sum_{i = 0}^{k - 1}\binom{k - 1}{i}\Gamma(k - i)\alpha^{m - i}\frac{\Gamma(m)\Gamma(i + 1)}{\Gamma(m + i + 1)} \\ = & c_1c_2\alpha^me^{-\alpha}\sum_{i = 0}^{k - 1}\frac{(k - 1)!(m - 1)!}{(m + i)!\alpha^i} = \alpha^me^{-\alpha}\sum_{i = 0}^{k - 1}\frac{1}{(m + i)!\alpha^i}. \end{align}

Now let's treat the numerator of $$(1)$$, careful calculation shows that \begin{align} & \int_0^\infty P[\xi > t, \eta < \alpha, \xi + \eta > \alpha] dt \\ = & \int_0^\infty \int_t^\infty P_\eta[\alpha - x < \eta < \alpha]f_\xi(x)dx dt \\ = & \int_0^\infty\int_t^\infty P[\eta < \alpha]f_\xi(x) dx dt - \int_0^\alpha\int_t^\alpha P[\eta < \alpha - x]f_\xi(x) dx dt \\ = & P[\eta < \alpha]E[\xi] - \int_0^\alpha\int_t^\alpha P[\eta < \alpha - x]f_\xi(x) dx dt \\ = & kP[\eta < \alpha] - \int_0^\alpha\int_t^\alpha P[\eta < \alpha - x]f_\xi(x) dx dt. \end{align} The latter term of the above integral can be expanded as: \begin{align} & c_1c_2\int_0^\alpha \int_t^\alpha\left[\int_0^{\alpha - x} e^{-y}y^{m - 1}dy\right]e^{-x}x^{k - 1}dxdt \\ = & c_1c_2\int_0^\alpha\left[\int_0^{\alpha - y}e^{-x}x^{k}dx\right]e^{-y}y^{m - 1}dy \\ = & c_1c_2\int_0^\alpha\left[\Gamma(k + 1) - \int_{\alpha - y}^\infty e^{-x}x^{k}dx\right]e^{-y}y^{m - 1}dy \\ = & c_1k!P[\eta < \alpha] - c_1c_2\int_0^\alpha\left[\int_{\alpha -y}^\infty e^{-x}x^{k}dx\right]e^{-y}y^{m - 1}dy \\ = & kP[\eta < \alpha] - c_1c_2\int_0^\alpha\left[\int_{\alpha -y}^\infty e^{-x}x^{k}dx\right]e^{-y}y^{m - 1}dy. \end{align} Therefore $$\int_0^\infty P[\xi > t, \eta < \alpha, \xi + \eta > \alpha] dt = c_1c_2\int_0^\alpha\left[\int_{\alpha -y}^\infty e^{-x}x^{k}dx\right]e^{-y}y^{m - 1}dy. \tag{3}$$

Fortunately, $$(3)$$ is almost the same as $$(2)$$, except for $$k$$ in $$(2)$$ will be replaced by $$k + 1$$ in $$(3)$$, whence a direct analogy gives $$\int_0^\infty P[\xi > t, \eta < \alpha, \xi + \eta > \alpha] dt = \alpha^me^{-\alpha}\sum_{i = 0}^{k}\frac{1}{(m + i)!\alpha^i}.$$

Finally, we can conclude that $$E[\xi | \eta < 1000, \xi + \eta > 1000] = \frac{\sum_{i = 0}^{k}\frac{1}{((n - 1)k + i)!1000^i}}{\sum_{i = 0}^{k - 1}\frac{1}{((n - 1)k + i)!1000^i}}.$$

• It might be a good idea to include a short summary of how one can evaluate (1), (2) and (3) to get a numeric solution to the problem, given that there appears to be no closed-form solution. – Ed P Jan 28 '19 at 4:04