Re: A probability problem about calculating the expectation

jing.hu@xxxxxxxxx wrote:
> Thank you very much for your inputs!
>
> The recorded value, also the Y you mentioned, does not need to be
> minimum of the realization of the random sequence which is above the T
> as well, but it is the first value above the T when scanning the
> sequence from X_1 to X_N. To clarify my point, the following is an
> example:
>
> Suppose N=4, lamda=1 or whatever (does not matter here, since the
> numbers are made purposely), T=1, a realization of the 4-tuple vector
> is {0.1, 1.2, 1.1, 0.9}. Then we pick the second one since it is above
> 1 at the first time from left to right and not neccessary to be the
> third one according to your definition of Y.
>

You really ought to quote a previous mesage. I now understand your
problem to be:

Let X1,..., Xn be i.i.d. Exponential(a) r.v.'s; let X0 = 0. Let K
be the smallest value in the union of {0} and {i: X_i >= t}. Let
Y = X_K. You want EY. Call this value y_n.

Let I be the indicator variable for {X1 >= t}. Then

y_n = EY = EE[Y | I] = (t + 1/a) exp(-at) + y_(n-1) [1 - exp(-at)],
and
y0 = 0.

Solve this first-order linear difference equation for y.

-SJH

.

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