2-dimensional random walk

Question

b = RandomVariate[NormalDistribution[0, 1]];
ListlinePlot[b, PlotRange -> 35]

I am so new to Mathematica. I am trying to generate a 2-dimensional walk with variance =1 and plot this. However, I do not get its plot.

Can you help me? Thank you so much.

cvgmt · Accepted Answer · 2021-01-26 02:01:18Z

6

FoldList[Plus, {0, 0}, RandomVariate[NormalDistribution[0, 1], {500, 2}]] // Line // Graphics

answered Jan 26, 2021 at 2:01

cvgmt

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kglr · Accepted Answer · 2021-01-26 18:59:41Z

7

Generate a list of random numbers and Accumulate:

b = Accumulate @ RandomVariate[NormalDistribution[0, 1], 500];

ListLinePlot[b, PlotRange -> 35]

For 2D, you can generate pairs of random numbers and Accumulate:

SeedRandom[1]

b2 = Accumulate @ RandomVariate[NormalDistribution[0, 1], {500, 2}];

ListLinePlot[b2, AspectRatio -> Automatic]

answered Jan 26, 2021 at 0:59

kglr

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2

$\begingroup$ Why do we need {500,2}? Thank you so much! $\endgroup$
– Bora
Jan 26, 2021 at 1:03
3

$\begingroup$ @Bora, RandomVariate[NormalDistribution[0, 1], {n, m}] generates an nXm matrix the entries of which are independent realizations of a random variable distributed NormalDistribution[0, 1]. $\endgroup$
– kglr
Jan 26, 2021 at 1:10
1

$\begingroup$ One last question. Is there any way to start the random walk at the origin? I am really grateful for your help. $\endgroup$
– Bora
Jan 26, 2021 at 1:49
2

$\begingroup$ Prepend the list with {0,0} before accumulating: b2 = Accumulate @ Prepend[RandomVariate[NormalDistribution[0, 1], {500, 2}], {0,0}] $\endgroup$
– kglr
Jan 26, 2021 at 2:15
1

$\begingroup$ Assuming that the x and y axes are equivalent, it's probably worth using the AspectRatio -> Automatic option on ListLinePlot. $\endgroup$
– Michael Seifert
Jan 26, 2021 at 18:57

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