in Probability
1,485 views
4 votes
4 votes
Consider the random process:

$X\left ( t \right )=U+Vt$

where $U$ is zero-mean Gaussian random variable and $V$ is a random variable uniformly distributed between $0$ and $2.$ Assume $U$ and $V$ statistically independent. The mean value of random process at $t=2$ is ___________
in Probability
by
1.5k views

4 Comments

@ankitgupta.1729

Is it in syllabus??:(

0
0
yeah, both gaussian and uniform distribution are in syllabus.
1
1

@ankitgupta.1729

what is $t=2$ means?

yes ans 2, elaborate now

0
0

1 Answer

2 votes
2 votes
$ X(2) = U + 2V $

$\Rightarrow E[X(2)] = E[ U + 2V] = E[X] + 2 E[V] $

Now, $E[X] = 0$ (given) and $ E[V] = \frac{0+2}{2} = 1$

$\Rightarrow E[X(2)] = 2$

3 Comments

havenot got this $ E[V] = \frac{0+2}{2} = 1$

how u finding mean??
0
0

For a random variable distributed uniformly between values $a$ and $b$, the mean is $\frac{a+b}{2}$.

Since $V$ is a random variable that is uniformly distributed between $0$ and $2$, It's mean ( or expected value ) = $ \frac{0+2}{2}= 1$

Refer to this link for its derivation: 

https://www.youtube.com/watch?v=hzb7u3lEtHA

0
0
Still not getting , how $E\left [ X\left ( 2 \right ) \right ]=2$
0
0

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true