# help with random variable convergence

#### omega

##### New Member
P(Yn=k)= ( (1-2^(-n-1))^(-1) )/ 2^(k+1) for k=0,1,2,...,n. The random variable Y has the geometric distribution where lambda is 1/2. How can I prove that {Yn} converges in distribution to Y?

Any help is appriciated.

I get fY(y)=1/2^(y+1)
but how do I prove the convergence?

#### BGM

##### TS Contributor
$$\Pr\{Y_n = k\} = \frac {1} {1 - 2^{-n-1}} \frac {1} {2^{k+1}}, k = 0, 1, 2, ..., n$$

Then the cumulative distribution of $$Y_n$$,

$$F_{Y_n}(y) = \Pr\{Y_n \leq y\} = \left\{\begin{array}{lll} 0 & \mbox{if} & y < 0 \\ {\displaystyle \frac {1} {1 - 2^{-n-1}} \left(1 - \frac {1} {2^{\lfloor y\rfloor + 1} } \right) }& \mbox{if} & 0 \leq y < n \\ 1 & \mbox{if} & y \geq n \end{array}\right .$$

$$\lim_{n\to +\infty} F_{Y_n}(y) = \left\{\begin{array}{lll} 0 & \mbox{if} & y < 0 \\ {\displaystyle 1 - \frac {1} {2^{\lfloor y\rfloor + 1} } }& \mbox{if} & y \geq 0 \end{array}\right . = F_Y(y)$$

#### omega

##### New Member
i'm a bit confused here, in your Fyn(y), for the equation where 0<=y<n, why did you put x(1-y) and not 1-(xy)? (x and y represents the parts of Yn)

Also in your Fyn(y), what about 1 if y>=1 since its a cdf ? if you don't add that part, doesn't it make it 1 if y>=0 ?

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