( 349 ) 



the coordinates y;, </ arc surpixssing the total length of the path. Then 

 the probability becomes a certainty and it follows that 



00 a: 



'Hi' 



sin pv sin q 



w 



= Ï Uv dw — ~ . — ^ -/{[/v' + lü') 







with the condition 



p and <7 ^ « -|- «1 + • • • + «n- !• 



In the general case of the rectangle the probability Wn (R) is 

 independent of q, as soon as its length is superior to that of the path. 



Assuming this to be the case, we remark that the value of the 

 slightly transformed integral 



4: r r sinpv simv . /< / to^\ 



W,(R) = -j jé.é. ^ . — ./ ((X "■ + 7J 







remains unaltered, when q increases indefinitely, and we conclude that 



00 00 00 



4 r sinpv C sinw 2 r sinpv 



Lim Wn{R) = — -J-f{v)dv . dw = - —£-f^v)dv. 



q=aa ^^J V J W St J V 







Thus we have solved another modification of Pearson's problem, 



1 

 for half the result, added to — , expresses the probability 



a 



00 



1 1 Csin pv 



2 St J V 







that the rambler, starting on his AA^alk at a distance p of a straight 

 frontier F, after Avalking along n stretches, will arrive at that side 

 of the frontier he came from ^). 



As before we are enabled in a particular case by the problem 

 itself to assign the A-alue of the integral. If we suppose that the 

 rambler cannot reach the frontier, that is, if we take 



P > « + «1 + • • • + ««-1 > 

 the probability becomes a certainty and we find 



1) Obviously the probability Wn (F) might have been derived from the proba- 

 bility TF;i-fi (w4-2? : wttfti . . . «n-i) by making « indefinitely large. Therefore we 

 may conclude that 



00 00 



^^^— , 11/^ sinvp 

 T^{na)-\-p)Jg{>ao) f{u)du = — --^ | f{v)dv. 

 2 Jtj V 



