# Thread: Quick question on exponential distribution

1. ## Quick question on exponential distribution

I just wanted to make sure I was doing these problems correctly, as I don't have the solutions to any of them, or previous work to base my answers on.

given an exponential distribution compute:

a) $E[X^2 -2X|X=1]$

what I was thinking it's simply $\int_1^\infty \lambda e^{-\lambda X} \cdot X^2 \ dX - 2\int_1^\infty \lambda e^{-\lambda X} \cdot X \ dX$

b) $P\{X>1|X<3\} = \int_1^{\infty} \lambda e^{-\lambda x} \ dx \cdot \bigg{(} \int_0^3 \lambda e^{-\lambda x} \ dx \bigg{)}^{-1}$

c) $P\{X_3>X_1+X_2\} = \int_{X_1=0}^{X_1=\infty} \int_{X_2=0}^{X_2=\infty}\int_{X_3=X_1+X_2}^{X_3=\ infty}\lambda_1 e^{-\lambda_1 X_1} \lambda_2 e^{-\lambda_2 X_2}$ $\lambda_3 e^{-\lambda_3 X_3} \ dX_3 \ dX_2 \ dX_1$

d) compute $P\{X_1=\min\{X_1, \ X_2\}\}$ with rates $\lambda_1$ and $\lambda_2$

based on my book the probabilistic minimum of is just $\frac{1}{\sum_{i=1}^n \lambda_i}$ (no proof was given) so in this case it would be $\frac{1}{\lambda_1+\lambda_2}$ also I would like to show that $\frac{1}{\sum_{i=1}^n \lambda_i}$ is the case, with 2 variable. I thought it would be:

$P\{X_1, < X_2\} + P\{X_2, < X_1\}$ but that give me 1 and

another way I considered was $P\{X_1, < X_2\} \cdot P\{X_2, < X_1\} = \frac{\lambda_1 \lambda_2}{(\lambda_1 +\lambda_2)^2}$

I finally considered $n[1-F(y)]^{n-1} \cdot f(y)$ but I don't know what $\lambda$ to pick since they are both different.

2. Originally Posted by lllll
I just wanted to make sure I was doing these problems correctly, as I don't have the solutions to any of them, or previous work to base my answers on.

given an exponential distribution compute:

a) $E[X^2 -2X|X=1]$

what I was thinking it's simply $\int_1^\infty \lambda e^{-\lambda X} \cdot X^2 \ dX - 2\int_1^\infty \lambda e^{-\lambda X} \cdot X \ dX$
Did you write the question correctly? Given your answer, it looks likes you were asked to compute $E[X^2-2X|X\geq 1]$. Anyway, your answer equals $E[(X^2-2X){\bf 1}_{(X\geq 1)}]$, i.e. you forgot to divide the result by $P(X\geq 1)$.

(By the way, it is usually advisable to write small $x$ in the integral because it is not exactly the same as $X$, without further precision)

b) $P\{X>1|X<3\} = \int_1^{\infty} \lambda e^{-\lambda x} \ dx \cdot \bigg{(} \int_0^3 \lambda e^{-\lambda x} \ dx \bigg{)}^{-1}$
Always start by writing what these conditional expectations stand for; in this case, it is $P(X>1|X<3)=\frac{P(X>1\mbox{ and }X<3)}{P(X<3)}$. Can you see why I said that?

c) $P\{X_3>X_1+X_2\} = \int_{X_1=0}^{X_1=\infty} \int_{X_2=0}^{X_2=\infty}\int_{X_3=X_1+X_2}^{X_3=\ infty}\lambda_1 e^{-\lambda_1 X_1} \lambda_2 e^{-\lambda_2 X_2}$ $\lambda_3 e^{-\lambda_3 X_3} \ dX_3 \ dX_2 \ dX_1$
OK, for a start.

d) compute $P\{X_1=\min\{X_1, \ X_2\}\}$ with rates $\lambda_1$ and $\lambda_2$

based on my book the probabilistic minimum of is just $\frac{1}{\sum_{i=1}^n \lambda_i}$ (no proof was given) so in this case it would be $\frac{1}{\lambda_1+\lambda_2}$ also I would like to show that $\frac{1}{\sum_{i=1}^n \lambda_i}$ is the case, with 2 variable. I thought it would be:

$P\{X_1, < X_2\} + P\{X_2, < X_1\}$ but that give me 1 and

another way I considered was $P\{X_1, < X_2\} \cdot P\{X_2, < X_1\} = \frac{\lambda_1 \lambda_2}{(\lambda_1 +\lambda_2)^2}$

I finally considered $n[1-F(y)]^{n-1} \cdot f(y)$ but I don't know what $\lambda$ to pick since they are both different.
I don't get it; you want $P\{X_1=\min\{X_1, \ X_2\}\}$. Isn't it just the same as $P\{X_1\leq X_2\}$?