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We draw $l$ samples from an exponential distribution with parameter $\lambda$. We know that summation of those samples, $Y = \sum_{i=1}^{l} X_i$ is less than a threshold, $t$. If we draw one more sample and add it to previous samples, $Z = Y + X_{l+1}$, what is the probability that the new summation of is greater than $t$? $P(Z>t|Y\leq t)$ ?

Samples are i.i.d. I know that sum of $l$ i.i.d. exponential random variables with parameter $\lambda$ has Erlang distribution with parameter $\lambda$ and shape $l$.

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HINT

You are asking what is $\mathbb{P}[X>t - Y|Y \le t]$ when $X,Y$ are independent with their distributions known? Note $$ \mathbb{P}[A|B] = \frac{\mathbb{P}[A \cap B]}{\mathbb{P}[B]} $$ and the denominator in your case is a 1-distribution problem, whereas the numerator is a double integral...

UPDATE

Further note that $$ \mathbb{P}[X>t - Y \cap Y \le t] = \int_{y=-\infty}^{y=t} \int_{x=t-y}^{x=\infty} f_{X,Y}(x,y)dxdy $$ and since $X$ and $Y$ are independent, $$ f_{X,Y}(x,y) = f_X(x) f_Y(y), $$ so the integral becomes $$ \begin{split} \mathbb{P}[X>t - Y \cap Y \le t] &= \int_{y=-\infty}^{y=t} f_Y(y) \left(\int_{x=t-y}^{x=\infty} f_X(x) dx\right)dy \\ &= \int_{y=-\infty}^{y=t} f_Y(y) \left(1 - F_X(t-y)\right)dy \\ &= F_Y(t) - \int_{y=-\infty}^{y=t} f_Y(y) F_X(t-y) dy \end{split} $$

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  • $\begingroup$ Thanks for the hint, but I just can't understand the part about double integral. Could you please explain a bit more? $\endgroup$ – Yashar Feb 27 '18 at 20:42
  • $\begingroup$ I mean how to derive distribution of [$A \cap B$] $\endgroup$ – Yashar Feb 27 '18 at 22:02
  • $\begingroup$ @Yashar see the update $\endgroup$ – gt6989b Feb 28 '18 at 3:57
  • $\begingroup$ Now I got it, thanks! $\endgroup$ – Yashar Mar 1 '18 at 1:01

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