The Study of Memory Part 1 – Short Term Memory

1. The Study of Memory Part 1 – Short Term Memory

2. Three questions: • Why did people originally believe in STM independent of LTM? • What do we think of those reasons now? • Do we need the STM construct?

3. Why did people originally believe in STM independent of LTM? • Because of STM – LTM differences in: • Patient data • Capacity • Duration • Type of code • Serial position effect • Mechanism of loss

4. Patient data.

5. Patient data Issue: is there a patient with a selective inability to add to LTM , with sparing of STM? If so, that selective impairment could be used in an argument for an independent STM. The most famous of all memory patients is HM.

6. HM (Scoville & Milner, 1957) • Surgery to relieve severe epilepsy, in 1953, at age 27. • bilateral excision of medial temporal lobe • after surgery, HM had profound anterograde amnesia. Capable of little if any new learning. • Some retrograde amnesia.

7. In HM’s words: "At this moment everything looks clear to me, but what happened just before? That's what worries me. It's like waking from a dream; I just don't remember".

8. Psychological studies by Brenda Milner. • HM has: • Good vocabulary and language; normal IQ • No attention disorder.

9. HM– does not know, for example: where he lives who cares for him what he ate at his last meal what year it is who the President of the United States is or how old he is. In 1982, he failed to recognize a picture of himself that had been taken on his 40th birthday in 1966.

10. Declarative tasks – asking HM what he knows: HM cannot learn (and later recall) new photographs of people verbal material sequences of digits complex geometric designs nonsense patterns. Cannot expand his digit span.

11. Procedural tasks – observing what HM can do. • Milner (1962) trained H. M. on a mirror-drawing task. • HM, like normal people, improves with practice. But he denies having practice. • Cohen and Corkin (1981) showed a similar result on the Tower of Hanoi puzzle.

12. HM – Conclusion: Though HM can learn procedures he cannot acquire new declarative learning. LTM impaired. But STM spared. Argument in favour of view that STM and LTM are independent.

13. Capacity

14. Capacity Issue: is STM different from LTM in capacity? If so, that supports view that LTM and STM are independent. Capacity of LTM is essentially infinite. What is capacity of STM?

15. Capacity • Shepard & Tehgtsoonian (1961) • Presented 200 3-digit numbers in a row. • E.g. … 492, 865, 931, 758… 865, … • Task: report when you hear a repeated number

16. Shepard & Teghtsoonian (1961) • I.V.: Interval between 1st and 2nd appearance • D.V.: Probability of noticing repetition • Repetition can only be noticed if first occurrence is still in memory. • Question: Are there separate forgetting functions for LTM and STM?

17. Shepard & Teghtsoonian (1961) • Result: • P(noticing repetition) fell dramatically at first • Steep decline ended at interval = 7 items • P(noticing) then fell more gradually, asymptoting at 60%

18. Shepard & Teghtsoonian (1961) – Interpretation: • Initial steep decline in P (noticing) occurs because response coming from STM. • More gradual occurs when response depends upon LTM. • Two forgetting functions – two memory stores, one large and one small.

19. Question: Why should STM have so small a capacity? Sensory memory has large capacity. LTM has large capacity. Why did we evolve a limited capacity store between two large capacity stores?

20. Answer: If STM was any larger, it would take too long to search through. When we need information from STM, to choose or guide a response, we need it fast. Things have to be processed fast in STM…

21. Duration

22. Duration. Issue: how long do STM traces last? LTM traces last a long time – possibly your whole life. If STM traces last less time, that supports the view that STM and LTM are independent.

23. Duration – how long does stuff last in STM? Brown (1958) and Peterson & Peterson (1959) Task: subjects see a simple stimulus (e.g., BRG) and have to report it back after an interval. Rehearsal is prevented by having them count backwards during retention interval. I.V. = length of interval in seconds.

24. % Correct as function of delay in Brown/Peterson task

25. Brown / Peterson: Result: after 18 seconds, subjects can no longer report stimulus. Interpretation: there is a memory system in which things must be rehearsed, or they are lost. But we don’t have to rehearse things in LTM – so there must be a second memory system – STM.

26. Type of code

27. Type of code. Issue: every stimulus has multiple aspects – e.g. colour brightness shape category name All are found in LTM. Which are found in STM?

28. Brown/Peterson task • Many similar studies reported in the ’60s. • Most errors phonological – e.g., P for T. • Errors based on shape were rare – e.g., C for O. • No semantic errors observed (or possible). • Conclusion: STM uses a phonological code.

29. Serial Position Effect

30. Serial Position Effect: In ordered recall, subjects recall a list in the order it was given. Out-of-order responses are counted as errors. Accuracy is higher for the beginning and end of the list, lower for the middle of the list.

31. % Position in list

32. Serial Position Effect Better performance at beginning of list is called Primacy Effect. Better performance at end of list is called Recency effect. Theory: Primacy produced by LTM. Recency produced by STM

33. Mechanism of Loss

34. How are things lost from memory – if at all? Decay? Interference? Retrieval failure? Originally, LTM loss was blamed on interference and STM loss was blamed on decay – as in Brown / Peterson paradgim.

35. 2. What do we think of those reasons now?

36. Differences between STM and LTM (taken in slightly different order this time): • Type of code • Serial position effect • Mechanism of loss • Patient data • Capacity • Duration • Do these reasons survive?

37. Type of code: • Original argument – any kind of code in LTM, only phonological codes in STM. • We now know – that STM can contain any kind of code. • See, for example, Brooks (1968), and Wickens Release from Proactive Inhibition studies.

38. Serial Position Effect: Original argument – Primacy effect produced by LTM, Recency effect produced by STM. We now know – that both Primacy and Recency effects can be found in pure LTM studies (e.g., recalling U.S. Presidents). Thus, recency effect cannot be taken as “empirical signature” of STM.

39. Mechanism of loss: Original argument – information lost from STM through decay, from LTM through interference. We now know – that information can be lost from STM through interference.

40. Duration • Original argument: • newly-acquired memories must be rehearsed to survive • but older memories do not need to be • therefore, new and old memories must be in separate stores.

41. Duration: • Alternative account: • traces in LTM are vulnerable until they have been consolidated. • new items are more vulnerable to loss than ‘established’ items.

42. HM: If traces in LTM are vulnerable until they have been consolidated, then HM’s problem is that he cannot consolidate. He has normal digit span because new items can be inserted in LTM. But he has anterograde amnesia because new items cannot be consolidated in LTM.

43. That leaves only Capacity … • Capacity is capacity of the Articulatory Loop – which is used for rehearsal of information and for planning articulation. • AL is not a short-term memory. • For example, you cannot search your articulatory loop, the way you can search memory.

44. Articulatory loop. Capacity is determined by rate of loss. You can rehearse about 7 items. If you try to rehearse more than 7 items, the first ones will be lost before you finish one cycle through the list and go back to the beginning.

45. Articulatory loop in action: Memory load = r l z t c j a Articulatory loop rehearses: r l z t c j a .. r l z t c j a .. r l z t c j a .. ‘r’ is still in loop when you finish ‘a.’

46. Articulatory loop in action: Memory load = r l z t c j a m k s c p y Articulatory loop rehearses: r l z t c j a m k s c p y .. ‘r’ is no longer in loop by the time you finish ‘y’ so cannot be rehearsed – is lost.

47. 3. Do we need the STM construct?

48. No. • We can explain all memory phenomena in terms of LTM and the articulatory loop. • All we need is two premises: • Limited capacity in articulatory loop. • Items in LTM are vulnerable to loss until they have been consolidated.